Chapter 15

Equations of Mathematical Physics

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Section 15.1Forming Second-Order Partial Differential Equations[20/0%]

Problem 5797

A perfectly flexible homogeneous string, under the influence of a strong tension $T_{0}$ , is in a rectilinear position of equilibrium.

Prove that the equation of small vibrations of the string has the form

\frac{\partial ^{2} u}{\partial x^{2}} = \frac{1}{a^{2}} \cdot \frac{\partial ^{2} u}{\partial t^{2}}, \quad a = \sqrt{\frac{T_{0}}{\rho }},

where $u$ is the displacement of the point with abscissa $x$ , $t$ --- time, and $\rho$ --- the linear density of the string (Fig. 15.1).

Fig. 15.1

Problem 5798

A thin homogeneous rod undergoes longitudinal vibrations in which its cross-sections, remaining planar, move along the axis $OX$ . Prove that the equation of small vibrations has the form

\frac{\partial ^{2} u}{\partial x^{2}} = \frac{1}{a^{2}} \cdot \frac{\partial ^{2} u}{\partial t^{2}}, \quad a = \sqrt{\frac{E}{\rho }},

where $u$ is the displacement of the cross-section with abscissa $x$ , $\rho$ --- the density, and $E$ --- the modulus of elasticity of the material of the rod (Fig. 15.2).

Fig. 15.2

Problem 5799

Prove that the equation of longitudinal vibrations of a conical rod has the form

\frac{\partial }{\partial x} \left(\left(1-\frac{x}{h}\right)^{2} \frac{\partial u}{\partial x}\right) = \frac{1}{a^{2}} \left(1-\frac{x}{h}\right)^{2} \frac{\partial ^{2} u}{\partial t^{2}}, \quad a = \sqrt{\frac{E}{\rho }},

where $u$ is the displacement of the rod with abscissa $x$ , and $h$ --- the height of the cone (Fig. 15.3).

Fig. 15.3

Problem 5800

Torsional vibrations of a rod are vibrations in which its cross-sections rotate relative to one another, each rotating about the axis of the rod. Prove that the equation of small vibrations of a round rod is

\frac{\partial ^{2} \theta }{\partial x^{2}} = \frac{1}{a^{2}} \cdot \frac{\partial ^{2} \theta }{\partial t}, \quad a = \sqrt{\frac{GI}{k}},

where $\theta$ is the angle of rotation of the cross-section with abscissa $x$ , $G$ --- the shear modulus, $I$ --- the polar moment of inertia of the cross-section, and $k$ --- the moment of inertia of a unit-length portion of the rod relative to the axis of the rod.

Problem 5801

A perfectly flexible homogeneous string is fastened at one end and, under the action of its own weight, is in a vertical position of equilibrium. Prove that the equation of small vibrations of the string has the form

\frac{\partial }{\partial x} \left((l-x) \frac{\partial u}{\partial x}\right) = \frac{1}{a^{2}} \cdot \frac{\partial ^{2} u}{\partial t^{2}}, \quad a = \sqrt{g},

where $u$ is the displacement of the point with abscissa $x$ , $l$ --- the length of the string, and $g$ --- the acceleration of gravity (Fig. 15.4).

Fig. 15.4

Problem 5802

A perfectly flexible string of length $l$ is suspended by the end $x = l$ , and to the other end $x = 0$ a weight of $P$ kg is attached. The linear density $\rho$ of the string varies according to the law

\rho = \frac{A}{\sqrt{l_{1}+x}},

where the constants $A$ and $l_{1}$ are related to the mass $M$ of the weight by

M = 2A\sqrt{l_{1}}.

Prove that the equation of small vibrations of the string about its equilibrium position has the form

\frac{\partial ^{2} u}{\partial \theta ^{2}} = \frac{1}{a^{2}} \cdot \frac{\partial ^{2} u}{\partial t^{2}}, \quad a = \sqrt{\frac{g}{2}}, \ \text{where} \ \theta = \sqrt{l+l_{1}}-\sqrt{l_{1}+x}.

Problem 5803

A heavy homogeneous string of length $l$ , fastened by its upper end ( $x = 0$ ) to a vertical axis, rotates about this axis with constant angular velocity $\omega$ (Fig. 15.5).

Prove that the equation of small vibrations of the string about its vertical equilibrium position has the form

\frac{\partial ^{2} u}{\partial t^{2}} = a^{2} \frac{\partial }{\partial x} \left(x \frac{\partial u}{\partial x}\right) + \omega ^{2} u,

where $a = \sqrt{g}$ .

Fig. 15.5

Problem 5804

A perfectly flexible homogeneous string is fastened at one end, and in equilibrium lies along the axis $x$ . Taken out of equilibrium, it rotates with constant angular velocity $\omega$ about the axis $u$ (Fig. 15.6).

Prove that the equation of small vibrations of the string about its relative equilibrium position has the form

\frac{\partial }{\partial x} \left((l^{2}-x^{2}) \frac{\partial u}{\partial x}\right) = \frac{1}{a^{2}} \cdot \frac{\partial ^{2} u}{\partial t^{2}}, \quad a = \frac{\omega }{\sqrt{2}},

where $l$ denotes the length of the string.

Fig. 15.6

Problem 5805

An inhomogeneous string, whose density varies according to the law

\rho = \frac{a}{\sqrt{b^{2}-x^{2}}}; \quad b > l,

is fastened by the end $x = 0$ to a fixed axis, while to the other end $x = l$ a ball is attached, whose mass is expressed by the equality

M = \frac{a}{l}\sqrt{b^{2}-l^{2}}.

Prove that when the string rotates with constant angular velocity $\omega$ about the indicated axis, the equation of small vibrations has the form

\frac{\partial ^{2} u}{\partial t^{2}} = \omega ^{2} \frac{\partial ^{2} u}{\partial y^{2}},

where

y = \arcsin \frac{x}{b}.

Problem 5806

Derive the equation of transverse vibrations of a string fastened at its ends, taking as the starting point Hamilton's principle of minimal action.

Remark.

Hint. For small vibrations of the string, the potential and kinetic energies are expressed by the formulas:

U = \frac{T_{0}}{2} \int _{0}^{l} \left(\frac{\partial u}{\partial x}\right)^{2} dx, \quad T = \frac{\rho }{2} \int _{0}^{l} \left(\frac{\partial u}{\partial t}\right)^{2} dx,

where $T_{0}$ is the magnitude of the initial tension, and $l$ --- the length of the string. According to Hamilton's principle, for any moments of time $t_{1}$ and $t_{2}$ the equality

\delta \int _{t_{1}}^{t_{2}} (U-T) \, dt = 0

holds.

Problem 5807

Starting from Hamilton's principle, solve problems 5801 and 5804 (problem 5801: a perfectly flexible homogeneous string is fastened at one end and, under the action of its own weight, is in a vertical position of equilibrium; it is required to derive the equation of small vibrations of the string about this position; problem 5804: a perfectly flexible homogeneous string is fastened at one end and in equilibrium lies along the axis $x$ , and, taken out of equilibrium, rotates with constant angular velocity $\omega$ about the axis $u$ ; it is required to derive the equation of small vibrations of the string about its relative equilibrium position).

Problem 5808

A flexible homogeneous string is stretched over a perfectly smooth sphere of radius $R$ and, in equilibrium, is directed along the equator. Prove that the equation of small vibrations of the string has the form

\frac{\partial ^{2} \theta }{\partial \varphi ^{2}}+\theta = \frac{1}{a^{2}} \cdot \frac{\partial ^{2} \theta }{\partial t^{2}}, \quad a = \frac{1}{R}\sqrt{\frac{T_{0}}{\rho }},

where $\theta$ and $\varphi$ denote the geographic coordinates of the point $M$ , $T_{0}$ is the initial tension, and $\rho$ --- the linear density of the string (Fig. 15.7).

Fig. 15.7

Remark.

Hint. One may use Hamilton's principle, taking into account the following expressions for the potential and kinetic energies:

U = \frac{RT_{0}}{2} \int _{0}^{a} \left(\left(\frac{\partial \theta }{\partial \varphi }\right)^{2}-\theta ^{2}\right) d\varphi , \quad T = \frac{R^{3}\rho }{2} \int _{0}^{a} \left(\frac{\partial \theta }{\partial t}\right)^{2} d\varphi ,

where $aR = l$ and $l$ denotes the length of the string.

Problem 5809

A flexible homogeneous string of length $l$ is fastened by one of its ends at the pole of a perfectly smooth sphere and, under the action of its own weight, lies along an arc of a great circle. Prove that the equation of small vibrations of the string has the form

\frac{\partial }{\partial \varphi } \left((\cos \varphi - \cos a) \frac{\partial \theta }{\partial \psi }\right)+(\cos \varphi - \cos a)\theta = \frac{1}{a^{2}} \frac{\partial ^{2} \theta }{\partial t^{2}}, \quad a = \sqrt{\frac{g}{R}},

where $g$ denotes the acceleration of gravity, and $R$ --- the radius of the sphere (Fig. 15.8).

Fig. 15.8

Remark.

Hint. The expressions for the potential and kinetic energies have the following form:

U = \frac{g\rho R^{2}}{2} \int _{0}^{a} (\cos \varphi - \cos a) \left(\left(\frac{\partial \theta }{\partial \varphi }\right)^{2}-\theta ^{2}\right) d\varphi ,

T = \frac{\rho R^{3}}{2} \int _{0}^{a} \left(\frac{\partial \theta }{\partial t}\right)^{2} d\varphi .

Problem 5810

Prove that for a wire with continuously distributed ohmic resistance $R$ , capacitance $C$ , inductance $L$ , and insulation leakage $G$ , the differential equations satisfied by the voltage $p$ and the current $i$ have the form

\frac{\partial p}{\partial x}+L\frac{\partial i}{\partial t}+Ri = 0, \quad \frac{\partial i}{\partial x}+C\frac{\partial p}{\partial t}+Gp = 0.

Remark.

Hint. The first equation is derived from the law of induction applied to a very small portion of the wire. The second equation is obtained from the condition of continuity of current in the wire.

Problem 5811

A perfectly flexible homogeneous film (membrane) is stretched over a frame $C$ . This position of the membrane is taken as the coordinate plane $XOY$ . Prove that the equation of small transverse vibrations of the membrane has the form

\frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}} = \frac{1}{a^{2}} \cdot \frac{\partial ^{2} u}{\partial t^{2}}, \quad a = \sqrt{\frac{T_{0}}{\sigma }},

where $T_{0}$ is the force of the initial tension, computed per unit of the contour $C$ , and $\sigma$ denotes the surface density of the membrane (Fig. 15.9).

Fig. 15.9

Problem 5812

A thin homogeneous rod is isolated from the external space and has at the point $M(x)$ a temperature $u(x,t)$ , changing with time. Prove that this temperature satisfies the equation

\frac{\partial ^{2} u}{\partial x^{2}} = \frac{1}{a^{2}} \cdot \frac{\partial u}{\partial t}, \quad a = \sqrt{\frac{k}{c\rho }},

where $c$ is the specific heat, $k$ --- the coefficient of internal thermal conductivity, and $\rho$ --- the density of the material of the rod.

Problem 5813

Prove that the equation of heat propagation in a thin insulated plate has the form

\frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}} = \frac{1}{a^{2}} \cdot \frac{\partial u}{\partial t},

where the value of the constant $a$ is the same as in the previous problem (problem 5812: $a = \sqrt{k/(c\rho )}$ , where $c$ is the specific heat, $k$ --- the coefficient of internal thermal conductivity, and $\rho$ --- the density of the material).

Problem 5814

Given a homogeneous ring with a very small cross-section (Fig. 15.10). Taking into account the loss of heat into the medium surrounding the ring due to radiation through the lateral surface, prove that the equation of heat propagation in the ring is

\frac{1}{R^{2}} \frac{\partial ^{2} u}{\partial \theta ^{2}}-bu = \frac{1}{a^{2}} \frac{\partial u}{\partial t}, \quad a = \sqrt{\frac{k}{c\rho }}, \quad b = \frac{hl}{\sigma k},

where $\theta$ denotes the length of the arc of the ring, and the constants entering the equation have the following meaning: $k$ , $h$ --- coefficients of internal and external thermal conductivity, $c$ , $\rho$ --- the specific heat and density of the material of the ring, $\sigma$ , $l$ --- the area and perimeter of the cross-section.

Fig. 15.10

Problem 5815

Prove that the temperature of a homogeneous conductor, subject to the action of a constant current, satisfies the differential equation

\frac{\partial u}{\partial t} = \frac{k}{c\rho } \frac{\partial ^{2} u}{\partial x^{2}} - \frac{hp}{c\rho \omega }(u-u_{0}) + \frac{0{,}24 i^{2} r}{c\rho \omega ^{2}},

where $i$ and $r$ denote the current strength and the resistivity of the conductor, $p$ and $\omega$ --- the perimeter and area of its cross-section, and $u_{0}$ --- the temperature of the surrounding medium; the meanings of the letters $k$ , $h$ , $c$ , $\rho$ are the same as in problem 5813 (problem 5813: $k$ --- the coefficient of internal thermal conductivity, $c$ --- the specific heat, $\rho$ --- the density of the material).

Problem 5816

Prove that the equation of radial heat propagation in a homogeneous insulated sphere has the form

\frac{\partial ^{2} v}{\partial r^{2}} = \frac{1}{a^{2}} \frac{\partial v}{\partial t}, \quad v(r,t) = ru(r,t), \quad a = \sqrt{\frac{k}{c\rho }},

where $u$ is the temperature of the sphere at a point at distance $r$ from its center; the meanings of the letters are the same as in problem 5814 (problem 5814: $k$ , $h$ --- coefficients of internal and external thermal conductivity, $c$ , $\rho$ --- the specific heat and density of the material, $\sigma$ , $l$ --- the area and perimeter of the cross-section).

Section 15.2Reduction of Second-Order Linear Equations to Canonical Form[7/100%]

Problem 5817

Reduce the following differential equation to canonical form:

\frac{\partial ^{2} u}{\partial x^{2}} - \frac{\partial ^{2} u}{\partial y^{2}} = 0.

Problem 5818

Reduce the following differential equation to canonical form:

x^{2}\frac{\partial ^{2} u}{\partial x^{2}}+2xy\frac{\partial ^{2} u}{\partial x \partial y}+y^{2}\frac{\partial ^{2} u}{\partial y^{2}} = 0.

Problem 5819

Reduce the following differential equation to canonical form:

y^{2}\frac{\partial ^{2} u}{\partial x^{2}}+x^{2}\frac{\partial ^{2} u}{\partial y^{2}} = 0.

Problem 5820

Reduce the following differential equation to canonical form:

x^{2}\frac{\partial ^{2} u}{\partial x^{2}}-y^{2}\frac{\partial ^{2} u}{\partial y^{2}} = 0.

Problem 5821

Reduce the following differential equation to canonical form:

(l-x)\frac{\partial ^{2} u}{\partial x^{2}} - \frac{\partial ^{2} u}{\partial y^{2}} - \frac{\partial u}{\partial x} = 0.

Problem 5822

Reduce the following differential equation to canonical form:

(l^{2}-x^{2})\frac{\partial ^{2} u}{\partial x^{2}} - \frac{\partial ^{2} u}{\partial y^{2}} - 2x\frac{\partial u}{\partial x} - \frac{1}{4}u = 0.

Problem 5823

Find the change of variables under which the equation

\frac{\partial ^{2} u}{\partial x \partial y} - \beta ' \ctg (x-y)\frac{\partial u}{\partial x} + \beta \ctg (x-y)\frac{\partial u}{\partial y} + \beta \beta ' u = 0

is transformed into the equation

\frac{\partial ^{2} v}{\partial \xi \partial \eta } - \frac{m}{\xi -\eta }\frac{\partial v}{\partial \xi } + \frac{n}{\xi -\eta }\frac{\partial v}{\partial \eta } + \frac{p}{(\xi -\eta )^{2}}v = 0.

Section 15.3The Method of Characteristics[17/24%]

Problem 5824

Prove that the general integral of the equation

\frac{\partial ^{2} u}{\partial x} = \frac{1}{a^{2}} \frac{\partial ^{2} u}{\partial t^{2}}

has the form $u = \varphi (x-at)+\psi (x+at)$ , where $\varphi$ and $\psi$ are arbitrary functions.

Problem 5825

Show that the general solution of the equation

\frac{\partial }{\partial x} \left(\left(1-\frac{x}{h}\right)^{2} \frac{\partial u}{\partial x}\right) = \frac{1}{a^{2}} \left(1-\frac{x}{h}\right)^{2} \frac{\partial ^{2} u}{\partial t^{2}}

can be given the form

u = \frac{\varphi (x-at)+\psi (x+at)}{x-h}.

Problem 5826

Prove that the solution of the equation

\frac{\partial ^{2} u}{\partial x^{2}} = \frac{1}{a^{2}} \frac{\partial ^{2} u}{\partial t^{2}},

satisfying the conditions

u\Big|_{t=0} = f(x), \quad \frac{\partial u}{\partial t}\Big|_{t=0} = F(x),

is given by the expression

u = \frac{f(x-at)+f(x+at)}{2} + \frac{1}{2a}\int _{x-at}^{x+at} F(z) \, dz.

Remark.

Hint. One should use the result of problem 5824.

Problem 5827

A segment of a homogeneous string, fastened at the ends $x = 0$ and $x = l$ , has at the initial moment of time the shape of a parabola, symmetric with respect to the perpendicular drawn through the midpoint of the segment $(0, l)$ . Determine the shape of the string at the moments of time $t = \dfrac {l}{2a}$ and $t = \dfrac {l}{a}$ , assuming that the initial velocities are zero.

Problem 5828

Prove that the equation given in problem 5825 (problem 5825: $\dfrac {\partial }{\partial x}\left(\left(1-\frac{x}{h}\right)^{2}\frac{\partial u}{\partial x}\right) = \frac{1}{a^{2}}\left(1-\frac{x}{h}\right)^{2}\frac{\partial^{2} u}{\partial t^{2}}$ , with general solution $u = \dfrac {\varphi (x-at)+\psi (x+at)}{x-h}$ ), under the conditions

u\Big|_{t=0} = f(x), \quad \frac{\partial u}{\partial t}\Big|_{t=0} = F(x)

has a solution of the form

u = \frac{(h-x+at)f(x-at)+(h-x-at)f(x+at)}{2(h-x)} + \frac{1}{2a}\int _{x-at}^{x+at} \frac{h-z}{h-x}F(z) \, dz.

Problem 5829

Find the integral of the equation

(p-x)\frac{\partial ^{2} u}{\partial x^{2}} - \frac{1}{2}\frac{\partial u}{\partial x} - \frac{\partial ^{2} u}{\partial y^{2}} = 0, \quad 0 < x < p

under the conditions:

u\Big|_{y=0} = f(x); \quad \frac{\partial u}{\partial y}\Big|_{y=0} = F(x).

Problem 5830

Integrate the equation

(p^{2}-x^{2})\frac{\partial ^{2} u}{\partial x^{2}} - x\frac{\partial u}{\partial x} - \frac{\partial ^{2} u}{\partial y^{2}} = 0, \quad 0 < x < p

under the conditions:

u\Big|_{y=0} = f(x); \quad \frac{\partial u}{\partial y}\Big|_{y=0} = F(x).

Problem 5831

Prove that the general solution of the equation

x^{2}\frac{\partial ^{2} u}{\partial x^{2}}+2xy\frac{\partial ^{2} u}{\partial x \partial y}+y^{2}\frac{\partial ^{2} u}{\partial y^{2}} = 0

has the form $u = \varphi \left(\dfrac {y}{x}\right)+y\psi \left(\dfrac {y}{x}\right)$ .

Remark.

Hint. See problem 5818 (problem 5818: $x^{2}\frac{\partial^{2} u}{\partial x^{2}}+2xy\frac{\partial^{2} u}{\partial x \partial y}+y^{2}\frac{\partial^{2} u}{\partial y^{2}} = 0$ is reduced to the canonical form $\frac{\partial^{2} u}{\partial \eta^{2}} = 0$ by the substitution $\xi = y/x$ , $\eta = y$ ).

Problem 5832

Show that the general solution of the equation

y^{2}\frac{\partial ^{2} u}{\partial y^{2}} - x^{2}\frac{\partial ^{2} u}{\partial x^{2}} = 0

can be given the form

u = \varphi (xy)+\sqrt{xy}\, \psi \left(\frac{y}{x}\right).

Remark.

Hint. See problem 5820 (problem 5820: $x^{2}\frac{\partial^{2} u}{\partial x^{2}}-y^{2}\frac{\partial^{2} u}{\partial y^{2}} = 0$ is reduced to the canonical form $\frac{\partial^{2} u}{\partial \xi \partial \eta }-\frac{1}{2\xi }\frac{\partial u}{\partial \eta } = 0$ by the substitution $\xi = xy$ , $\eta = y/x$ ).

Problem 5833

Prove that the equation given in the preceding problem (problem 5832: $y^{2}\frac{\partial^{2} u}{\partial y^{2}} - x^{2}\frac{\partial^{2} u}{\partial x^{2}} = 0$ , with general solution $u = \varphi (xy)+\sqrt{xy}\, \psi (y/x)$ ), under the conditions

u\Big|_{y=1} = f(x), \quad \frac{\partial u}{\partial y}\Big|_{y=1} = F(x)

has the following solution:

u(x,y) = \frac{1}{2}f(xy)+\frac{y}{2}f\left(\frac{x}{y}\right) + \frac{1}{4}\sqrt{xy}\int _{xy}^{x/y} \frac{f(x)}{x^{3/2}} \, dx - \frac{1}{2}\sqrt{xy}\int _{xy}^{x/y} \frac{F(x)}{x^{3/2}} \, dx.

Remark.

Hint. One should use the result of the preceding problem.

Problem 5834

Prove that if, at $x = ct$ , $c = \text{const}$ , the equalities $u = f(t)$ , $\dfrac {\partial u}{\partial x} = F(t)$ hold, then the solution of the equation

\frac{\partial ^{2} u}{\partial x^{2}} = \frac{1}{a^{2}} \frac{\partial ^{2} u}{\partial t^{2}}

is given by the expression

u = \frac{1+\beta }{2}f\left(\frac{at+x}{a(1+\beta )}\right) + \frac{1-\beta }{2}f\left(\frac{at-x}{a(1-\beta )}\right) + \frac{a(1-\beta ^{2})}{2}\int _{\frac{t-\frac{x}{a}}{1-\beta }}^{\frac{t+\frac{x}{a}}{1+\beta }} F(z) \, dz,

where $\beta = \dfrac {c}{a}$ .

Remark.

Hint. One should use the result of problem 5824 and take into account the equalities

\frac{\partial u}{\partial t}\Big|_{x=ct} = \frac{\partial u}{\partial t}\Big|_{x=ct} + \frac{\partial u}{\partial x}\Big|_{x=ct} \cdot \frac{\partial x}{\partial t}\Big|_{x=ct}.

Problem 5835

At the end $x = 0$ of a cylindrical rod, long enough that it can be considered as extending in one direction to infinity, a disturbing harmonic force $A\sin \omega t$ acts. Prove that the displacement of the cross-section of the rod with abscissa $x$ is expressed by the formula

u = \begin{cases} 0 & \text{for } at \leqslant x; \\ A\sin \dfrac {\omega }{a}(at-x) & \text{for } at \geqslant x. \end{cases}

Remark.

Hint. Apply the method of characteristics to the integration of the equation

\frac{\partial ^{2} u}{\partial t^{2}} = a^{2}\frac{\partial ^{2} u}{\partial x^{2}}

under the boundary condition $u\big|_{x=0} = A\sin \omega t$ and under the initial conditions: $\dfrac {\partial u}{\partial x}\Big|_{t=0} = 0$ ; $\dfrac {\partial u}{\partial t}\Big|_{t=0} = 0$ , which hold for every positive $x$ .

Problem 5836

An ideal gas, whose density varies according to the law $\rho = f(at-x)$ , undergoes longitudinal vibrations in an unbounded cylindrical tube. Prove that the velocity of the gas in the cross-section with abscissa $x$ is expressed by the formula

v = \frac{af(at-x)+(v_{0}-a)f(at)}{f(at-x)},

where $v_{0}$ --- the velocity in the cross-section $x = 0$ .

Remark.

Hint. Use the continuity equation.

Problem 5837

An infinite string, in a rectilinear position of equilibrium, receives at the initial moment of time ( $t = 0$ ) a blow from a small hammer, whose mass equals $M$ , this hammer touching the string at the point $x = 0$ and having initial velocity $v_{0}$ .

Prove that at the moment of time $t > 0$ the disturbed string has the shape shown in figure 15.11, where $u_{1}$ --- the forward wave:

Fig. 15.11

u_{1} = \begin{cases} \dfrac {Mav_{0}}{2T_{0}}\left(1-e^{\frac{2T_{0}}{Ma^{2}}(x-at)}\right), & \text{for } x-at < 0; \\ 0, & \text{for } x-at > 0, \end{cases}

and $u_{2}$ --- the backward wave:

u_{2} = \begin{cases} \dfrac {MAv_{0}}{2T_{0}}\left(1-e^{-\frac{2T_{0}}{Ma^{2}}(x+at)}\right), & \text{for } x+at > 0; \\ 0, & \text{for } x+at < 0. \end{cases}

For the values of $a$ and $T_{0}$ see problem 5797 (problem 5797: $a = \sqrt{T_{0}/\rho }$ , where $T_{0}$ --- the tension, and $\rho$ --- the linear density of the string).

Remark.

Hint. When integrating the equation

\frac{\partial ^{2} u}{\partial t^{2}} = a^{2}\frac{\partial ^{2} u}{\partial x^{2}}

one should take into account the condition

M\frac{\partial ^{2} u_{1}}{\partial t^{2}}\Big|_{x=0} = M\frac{\partial ^{2} u_{2}}{\partial t^{2}}\Big|_{x=0} = -T_{0}\frac{\partial u_{2}}{\partial x_{2}}\Big|_{x=0} + T_{0}\frac{\partial u_{1}}{\partial x}\Big|_{x=0}.

Problem 5838

The origin of coordinates $x = 0$ is placed at the plane of separation of two gases (I) and (II), filling a very long cylindrical tube. The condensation of the gases is denoted by $s_{1}$ and $s_{2}$ . From region (I) a forward wave $s_{1} = F\left(t-\dfrac {x}{a_{1}}\right)$ moves, undergoing reflection and absorption at the cross-section $x = 0$ .

Prove that between the incident wave $F$ , the reflected wave $f$ , and the absorbed wave $\varphi$ there is a relation

\frac{f}{F} = \frac{a_{1}-a_{2}}{a_{1}+a_{2}}, \quad \frac{\varphi }{F} = \frac{2a_{2}}{a_{1}+a_{2}},

where $a_{2}$ denotes the speed of the absorbed wave.

Problem 5839

The region of initial vibrations of a homogeneous gas is a sphere of radius $R$ . The initial velocities of the gas particles are everywhere zero, and the initial density $S_{0}$ is constant inside the sphere and equal to zero outside it. Find the expression for the density $S$ at the moment of time $t > 0$ .

Remark.

Hint. Apply the method of characteristics to the integration of the equation

\frac{\partial ^{2}\left(r\Phi (r,t)\right)}{\partial t^{2}} = a^{2}\frac{\partial ^{2}\left(r\Phi (r,t)\right)}{\partial r^{2}}, \quad \left(S = \frac{1}{a^{2}}\frac{\partial \Phi }{\partial t}\right)

under the initial conditions

\Phi \big|_{t=0} = f(r) = 0; \quad \frac{\partial \Phi }{\partial t}\Big|_{t=0} = F(r) = \begin{cases} a^{2}S_{0} & \text{for } r < R; \\ 0 & \text{for } r > R. \end{cases}

Problem 5840

Solve the preceding problem (problem 5839), applying the Poisson integral:

\Phi (r,t) = \frac{1}{4\pi a}\left(\frac{\partial }{\partial t} \iint _{\Sigma } \frac{f(\alpha ,\beta ,\gamma )}{r} \, d\sigma + \iint _{\Sigma } \frac{F(\alpha ,\beta ,\gamma )}{t} \, d\sigma \right),

where $\Sigma$ denotes the sphere described with radius $r = at$ from the point $M$ , and $\alpha , \beta , \gamma$ are the coordinates of a variable point of the sphere $\Sigma$ (Fig. 15.12).

Fig. 15.12

Section 15.4The Riemann Method[5/40%]

Problem 5841

Integrate by the Riemann method the equation $\dfrac {\partial^{2} u}{\partial x^{2}} - \dfrac {\partial^{2} u}{\partial y^{2}} = 0$ under the conditions

u\Big|_{y=0} = f(x), \quad \frac{\partial u}{\partial y}\Big|_{y=0} = F(x).

Remark.

Hint. In this case the Riemann function equals unity.

Problem 5842

Solve by the Riemann method problem 5833 (problem 5833: prove that the equation $y^{2}\frac{\partial^{2} u}{\partial y^{2}} - x^{2}\frac{\partial^{2} u}{\partial x^{2}} = 0$ under the conditions $u\big|_{y=1} = f(x)$ , $\frac{\partial u}{\partial y}\big|_{y=1} = F(x)$ has the solution $u(x,y) = \frac{1}{2}f(xy)+\frac{y}{2}f(x/y) + \frac{1}{4}\sqrt{xy}\int_{xy}^{x/y} f(x)x^{-3/2}\, dx - \frac{1}{2}\sqrt{xy}\int_{xy}^{x/y} F(x)x^{-3/2}\, dx$ ).

Remark.

Hint. As the region of integration one should take the curvilinear triangle formed by the characteristics $MP$ , $MQ$ and the arc of the equilateral hyperbola $xy = 1$ (Fig. 15.13). The Riemann function is determined by the equality:

v(x,y;x_{0},y_{0}) = \sqrt{\frac{x_{0}}{x}}.

Fig. 15.13

Problem 5843

Show, using the Riemann method, that the integral of the equation $\dfrac {\partial^{2} \theta }{\partial \varphi^{2}}+\theta = \dfrac {1}{a^{2}}\dfrac {\partial^{2} \theta }{\partial t^{2}}$ , satisfying the conditions $\theta \big|_{t=0} = f(\varphi )$ ; $\dfrac {\partial \theta }{\partial t}\Big|_{t=0} = F(\varphi )$ , has the form

\theta (\varphi ,t) = \frac{f(\varphi -at)+f(\varphi +at)}{2} + \frac{1}{2}\int _{\varphi -at}^{\varphi +at} \Phi (\varphi ,t,z) \, dz,

where

\Phi (\varphi ,t,z) = \frac{1}{a}F(z)J_{0}\left(\sqrt{(z-\varphi )^{2}-a^{2}t^{2}}\right) - atf(z)\frac{J_{0}'\left(\sqrt{(z-\varphi )^{2}-a^{2}t^{2}}\right)}{\sqrt{(z-\varphi )^{2}-a^{2}t^{2}}}.

Remark.

Hint. The Riemann function is determined by the equality

v(x,y;x_{0},y_{0}) = J_{0}\left(\sqrt{(x-x_{0})(y-y_{0})}\right),

where $J_{0}(x)$ denotes the Bessel function of order zero.

Problem 5844

Applying the Riemann method, prove that, under the conditions $u\big|_{t=0} = f(x)$ , $\dfrac {\partial u}{\partial t}\Big|_{t=0} = \varphi (x)$ , the equation

(l-x)\frac{\partial ^{2} u}{\partial x^{2}} - \frac{\partial u}{\partial x} = \frac{1}{a^{2}}\frac{\partial ^{2} u}{\partial t^{2}}

has the solution

u(x,t) = \frac{\sqrt{\sqrt{l-x}-\frac{at}{2}}\, f\left(x+\sqrt{l-x}\, at-\frac{a^{2}t^{2}}{4}\right)}{2\sqrt[4]{l-x}} + \frac{\sqrt{\sqrt{l-x}+\frac{at}{2}}\, f\left(x-\sqrt{l-x}\, at-\frac{a^{2}t^{2}}{4}\right)}{2\sqrt[4]{l-x}} + \frac{1}{\sqrt[4]{l-x}}\int _{\sqrt{l-x}-\frac{at}{2}}^{\sqrt{l-x}+\frac{at}{2}} \Phi (x,t,z) \, dz;

\Phi (x,t,z) = \frac{\sqrt{z}}{a}\varphi (l-z^{2})F\left(\frac{1}{2},\frac{1}{2},1,\frac{\frac{a^{2}t^{2}}{4}-(z-\sqrt{l-x})^{2}}{4\sqrt{l-x}\, z}\right) + \frac{atf(l-z^{2})}{8\sqrt{l-x}\sqrt{z}}F'\left(\frac{1}{2},\frac{1}{2},1,\frac{\frac{a^{2}t^{2}}{4}-(z-\sqrt{l-x})^{2}}{4\sqrt{l-x}\, z}\right),

where $F(\alpha ,\beta ,\gamma ,x) = 1+\dfrac {\alpha \cdot \beta }{1\cdot \gamma }x+\dfrac {\alpha (\alpha +1)\beta (\beta +1)}{1\cdot 2\cdot \gamma (\gamma +1)}x^{2}+\ldots$ is the hypergeometric series.

Remark.

Hint. The given equation must be reduced to the simplest canonical form (see problem 5820). As the region of integration one should take the triangle formed by the characteristics $MP$ , $MQ$ and the bisector $y = -x$ (Fig. 15.14). The Riemann function is determined by the equality

v(x,y;x_{0},y_{0}) = F\left(\frac{1}{2},\frac{1}{2},1,\frac{(x-x_{0})(y-y_{0})}{(x_{0}-y_{0})(x-y)}\right).

Fig. 15.14

Problem 5845

Integrate by the Riemann method the equation

(l^{2}-x^{2})\frac{\partial ^{2} u}{\partial x^{2}} - 2x\frac{\partial u}{\partial x} - \frac{\partial ^{2} u}{\partial y^{2}} - \frac{1}{4}u = 0, \quad 0 < x < l,

subjecting its integral to the conditions

u\Big|_{y=0} = f(x), \quad \frac{\partial u}{\partial y}\Big|_{y=0} = \varphi (x).

Remark.

Hint. The Riemann function has the following form:

v(x,y;x_{0},y_{0}) = F\left(\frac{1}{2},\frac{1}{2},1,\frac{\sin (x-x_{0})\sin (y-y_{0})}{\sin (x_{0}-y_{0})\sin (x-y)}\right).

Section 15.5The Fourier Method[54/83%]

Problem 5846

For a homogeneous string of length $l$ , fastened at the ends $x = 0$ and $x = l$ , the deviation from the rectilinear equilibrium position and the velocities of the points are known at the initial moment of time. Study by the Fourier method the transverse vibrations of the string (the equation of small vibrations of the string --- problem 5797: $\frac{\partial^{2} u}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial^{2} u}{\partial t^{2}}$ , $a = \sqrt{T_{0}/\rho }$ ).

Remark.

Hint. The problem reduces to integrating the equation

\frac{\partial ^{2} u}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial ^{2} u}{\partial t^{2}}, \quad a = \sqrt{\frac{T_{0}}{\rho }}

and to satisfying the conditions: $u\big|_{x=0} = 0$ , $u\big|_{x=l} = 0$ for every $t$ ; $u\big|_{t=0} = f(x)$ , $\dfrac {\partial u}{\partial t}\Big|_{t=0} = F(x)$ for $0 < x < l$ .

Problem 5847

Study by means of the Fourier method the longitudinal vibrations of a cylindrical rod of length $l$ , one end of which is fixed while the other is free (the equation of small longitudinal vibrations of the rod --- problem 5798: $\frac{\partial^{2} u}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial^{2} u}{\partial t^{2}}$ , $a = \sqrt{E/\rho }$ ).

Remark.

Hint. The problem reduces to integrating the equation

\frac{\partial ^{2} u}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial ^{2} u}{\partial t^{2}}, \quad a = \sqrt{\frac{E}{\rho }}

under the conditions: $u\big|_{x=0} = 0$ , $\dfrac {\partial u}{\partial x}\Big|_{x=l} = 0$ for every $t$ ; $u\big|_{t=0} = f(x)$ , $\dfrac {\partial u}{\partial t}\Big|_{t=0} = F(x)$ for $0 < x < l$ .

Problem 5848

Solve the preceding problem (problem 5847) under the condition that both ends of the rod are free.

Problem 5849

Integrate by the Fourier method the equation of torsional vibrations of a shaft $\dfrac {\partial^{2} \theta }{\partial x^{2}} = \dfrac {1}{a^{2}}\dfrac {\partial^{2} \theta }{\partial t^{2}}$ , $a = \sqrt{GI/k}$ , under the condition that one of the ends of the shaft is fixed, and to the other a pulley is attached (see problem 5801).

Remark.

Hint. The integral of the vibration equation satisfies the conditions

\theta \big|_{x=0} = 0, \quad \frac{\partial ^{2} \theta }{\partial t^{2}}\Big|_{x=l} = -c^{2}\frac{\partial \theta }{\partial x}\Big|_{x=l}

for every $t$ , and $\theta \big|_{t=0} = f(x)$ , $\dfrac {\partial \theta }{\partial t}\Big|_{t=0} = F(x)$ for $0 < x < l$ , where $c = \sqrt{GI/k_{1}}$ and $k_{1}$ denotes the moment of inertia of the pulley relative to the axis of the shaft.

Problem 5850

Solve the same problem (problem 5849), assuming that two identical pulleys are attached to the ends of the shaft.

Remark.

Hint. The boundary conditions have the form

\frac{\partial ^{2} \theta }{\partial t^{2}}\Big|_{x=0} = c^{2}\frac{\partial \theta }{\partial x}\Big|_{x=0}; \quad \frac{\partial ^{2} \theta }{\partial t^{2}}\Big|_{x=l} = -c^{2}\frac{\partial \theta }{\partial x}\Big|_{x=l}, \quad \text{where} \quad c = \sqrt{\frac{GI}{k_{1}}}.

Problem 5851

To the end $x = l$ of a homogeneous rod, fastened at the point $x = 0$ , a weight of $P$ kg is suspended. Study the forced longitudinal vibrations of the rod, assuming that it is acted on by an external continuously distributed force $\rho Y(x,t)$ .

Remark.

Hint. The problem reduces to integrating the equation

\frac{\partial ^{2} u}{\partial t^{2}} = a^{2}\frac{\partial ^{2} u}{\partial x^{2}}+Y(x,t), \quad a = \sqrt{\frac{E}{\rho }}

under the conditions

u\big|_{x=0} = 0; \quad \frac{\partial ^{2} u}{\partial t^{2}}\Big|_{x=l} = -c^{2}\frac{\partial u}{\partial x}\Big|_{x=l}, \quad c = \sqrt{\frac{gE}{P}};

u\big|_{t=0} = 0; \quad \frac{\partial u}{\partial t}\Big|_{t=0} = 0;

$g$ --- the acceleration of free fall.

Problem 5852

A string of length $l$ is fastened at the ends $x = 0$ and $x = l$ and at the initial moment of time is drawn aside at the point $x = c$ by a distance $h$ from the axis $OX$ ; it is then released without imparting any initial velocity to its points (Fig. 15.15). Determine the deviation $u(x,t)$ of the string for any moment of time (see problem 5846).

Fig. 15.15

Problem 5853

A string of length $l$ is fastened at the ends $x = 0$ and $x = l$ and at the initial moment of time, at the point $x = c$ , receives a blow from a small hammer (a piano string). It is assumed that during the blow the hammer touches a very small portion of the string, and that the blow itself lasts very briefly compared to the period of vibration of the string. Find the expression for the deviation $u(x,t)$ of the string (see problem 5846).

Remark.

Hint. The initial conditions of vibration: $f(x) = 0$ throughout the interval $(0,l)$ , $F(x) = 0$ throughout the interval $(0,l)$ except for a very small portion $(c-\varepsilon , c+\varepsilon )$ , where $F(x) > 0$ .

Problem 5854

A perfectly insulated electric wire of length $l$ , at a certain moment of time, taken as the initial one, is charged to a certain constant potential $p_{0}$ . Determine the magnitude of the voltage and the current $i$ after a time $t$ from the initial moment, assuming that one of the ends of the wire is short-circuited and the other is open (the equations of the line --- problem 5810: $\frac{\partial p}{\partial x}+L\frac{\partial i}{\partial t}+Ri = 0$ , $\frac{\partial i}{\partial x}+C\frac{\partial p}{\partial t}+Gp = 0$ ).

Remark.

Hint. The problem reduces to integrating the system of equations

\frac{\partial p}{\partial x}+L\frac{\partial i}{\partial t}+Ri = 0, \quad \frac{\partial i}{\partial x}+C\frac{\partial p}{\partial t} = 0

under the conditions: $p\big|_{x=0} = 0$ , $i\big|_{x=l} = 0$ for every $t$ , $p\big|_{t=0} = p_{0}$ , $i\big|_{t=0} = 0$ for $0 < x < l$ .

Problem 5855

A homogeneous string of length $l$ , under the influence of a strong tension $T_{0}$ , is in a rectilinear position of equilibrium and is fastened at the ends $x = 0$ and $x = l$ . At the point $x = c$ a concentrated harmonic force $\rho P(t) = \rho A\sin \omega t$ acts on the string.

Determine the deviation $u(x,t)$ of the string from the equilibrium position (Fig. 15.16).

Fig. 15.16

Remark.

Hint. The problem can be reduced to integrating the equations

\frac{\partial ^{2} u_{1}}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial ^{2} u_{1}}{\partial t^{2}}, \quad \frac{\partial ^{2} u_{2}}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial ^{2} u_{2}}{\partial t^{2}}, \quad a = \sqrt{\frac{T_{0}}{\rho }}

and to satisfying the boundary conditions:

u_{1}\big|_{x=0} = 0, \quad u_{1}\big|_{x=c} = u_{2}\big|_{x=c}, \quad u_{2}\big|_{x=l} = 0.

T_{0}\frac{\partial u_{2}}{\partial x}\Big|_{x=c} - T_{0}\frac{\partial u_{1}}{\partial x}\Big|_{x=c} + P\rho = 0.

Problem 5856

A homogeneous string of length $l$ , fastened at the ends $x = 0$ and $x = l$ , vibrates under the action of an external disturbing force $\rho Y(x,t)$ , distributed continuously along the entire length of the string. Find the deviations of the string from the equilibrium position, taking into account its free vibrations, caused by the initial deviation and the initial velocity.

Remark.

Hint. The equation of forced vibrations has the form

\frac{\partial ^{2} u}{\partial t^{2}} = a^{2}\frac{\partial ^{2} u}{\partial x^{2}}+Y(x,t); \quad a = \sqrt{\frac{T_{0}}{\rho }};

the integral of this equation is written as the sum $u = u_{1}+u_{2}$ , where $u_{1}$ is the integral of the homogeneous equation $\dfrac {\partial^{2} u}{\partial t^{2}} = a^{2}\dfrac {\partial^{2} u}{\partial x^{2}}$ , satisfying the conditions:

u_{1}\big|_{x=0} = 0, \quad u_{1}\big|_{x=l} = 0, \quad u_{1}\big|_{t=0} = f(x), \quad \frac{\partial u_{1}}{\partial t}\Big|_{t=0} = F(x),

and $u_{2}$ is the integral of the inhomogeneous equation, subject to the conditions:

u_{2}\big|_{x=0} = 0, \quad u_{2}\big|_{x=l} = 0, \quad u_{2}\big|_{t=0} = 0, \quad \frac{\partial u_{2}}{\partial t}\Big|_{t=0} = 0.

Problem 5857

Study the forced transverse vibrations of a string, fastened at the end $x = 0$ and subject at the end $x = l$ to the action of a disturbing harmonic force, causing a displacement equal to $A\sin \omega t$ .

Problem 5858

Solve the preceding problem (problem 5857) under the assumption that the string is fastened at the ends $x = 0$ and $x = l$ and that the external disturbing force is concentrated at the point $x = c$ of the string. Investigate the special case in which the force acts harmonically.

Remark.

Hint. The force concentrated at the point $x = c$ can be regarded as the limiting case of a force $\rho Y(x,t) > 0$ , continuously distributed over a very small portion $(c-\varepsilon , c+\varepsilon )$ .

Problem 5859

Two dissimilar strings of length $l_{1}$ and $l_{2}$ are joined together at the point $x = l_{1}$ and fastened at the ends $x = 0$ and $x = l_{1}+l_{2}$ . Determine the period of transverse vibrations.

Remark.

Hint. The problem reduces to integrating the equations

\frac{\partial ^{2} u_{1}}{\partial t^{2}} = a_{1}^{2}\frac{\partial ^{2} u_{1}}{\partial x^{2}}; \quad \frac{\partial ^{2} u_{2}}{\partial t^{2}} = a_{2}^{2}\frac{\partial ^{2} u_{2}}{\partial x^{2}}; \quad a_{1} = \sqrt{\frac{T_{0}}{\rho _{1}}}; \quad a_{2} = \sqrt{\frac{T_{0}}{\rho _{2}}}

under the conditions

u_{1}\big|_{x=0} = 0, \quad u_{2}\big|_{x=l_{1}+l_{2}} = 0, \quad u_{1}\big|_{x=l_{1}} = u_{2}\big|_{x=l_{1}};

\frac{\partial u_{1}}{\partial x}\Big|_{x=l_{1}} = \frac{\partial u_{2}}{\partial x}\Big|_{x=l_{1}}.

Problem 5860

A string, fastened at the ends, consists of two inhomogeneous parts $AB = l_{1}$ and $BC = l_{2}$ , with a ball of mass $M$ placed at the point $B$ . Prove that the period $T$ of transverse vibrations is determined from the equation

\frac{1}{a_{1}}\ctg \frac{2\pi l_{1}}{Ta_{1}} + \frac{1}{a_{2}}\ctg \frac{2\pi l_{2}}{Ta_{2}} = M\frac{2\pi }{TT_{0}},

where $T_{0}$ --- the magnitude of the tension of the string.

Remark.

Hint. One should take into account the condition

M\frac{\partial ^{2} u_{1}}{\partial t^{2}}\Big|_{x=l_{1}} = M\frac{\partial ^{2} u_{2}}{\partial t^{2}}\Big|_{x=l_{1}} = -T_{0}\frac{\partial u_{1}}{\partial x}\Big|_{x=l_{1}} + T_{0}\frac{\partial u_{2}}{\partial x}\Big|_{x=l_{1}}.

Problem 5861

A rod $AC$ , fastened at the beginning $A$ and free at the end $C$ , consists of two inhomogeneous parts $AB = l_{1}$ and $BC = l_{2}$ . Prove that the period $T$ of longitudinal vibrations of the rod is determined by the equation

\tg \frac{2\pi l_{1}}{a_{1}T} \cdot \tg \frac{2\pi l_{2}}{a_{2}T} = \frac{E_{1}a_{2}}{E_{2}a_{1}},

where $a_{1} = \sqrt{\dfrac {E_{1}}{\rho_{1}}}$ , $a_{2} = \sqrt{\dfrac {E_{2}}{\rho_{2}}}$ .

Remark.

Hint. Single out vibrations of one and the same frequency, satisfying the equations

\frac{\partial ^{2} u_{1}}{\partial t^{2}} = a_{1}^{2}\frac{\partial ^{2} u_{1}}{\partial x^{2}}; \qquad \frac{\partial ^{2} u_{2}}{\partial t^{2}} = a_{2}^{2}\frac{\partial ^{2} u_{2}}{\partial x^{2}}

under the conditions

u_{1}\big|_{x=l_{1}} = u_{2}\big|_{x=l_{1}}; \qquad E_{1}\frac{\partial u_{1}}{\partial x}\Big|_{x=l_{1}} = E_{2}\frac{\partial u_{2}}{\partial x}\Big|_{x=l_{1}}.

Problem 5862

A heavy extensible rod, whose length in the unstretched state equals $l$ , is suspended by the end $x = 0$ , while the end $x = l$ is left free. Study the forced vibrations of the rod.

Remark.

Hint. The problem reduces to integrating the equation

\frac{\partial ^{2} u}{\partial t^{2}} = a^{2}\frac{\partial ^{2} u}{\partial x^{2}}+g

under the conditions: $u\big|_{t=0} = 0$ ; $\dfrac {\partial u}{\partial t}\Big|_{t=0} = 0$ , $u\big|_{x=0} = 0$ ; $\dfrac {\partial u}{\partial x}\Big|_{x=0} = 0$ .

Problem 5863

Prove that the radial vibrations of a homogeneous gas, enclosed between two fixed concentric spheres of radii $R_{1}$ and $R_{2}$ , $R_{1} < R_{2}$ , can occur harmonically, with the period of the tone expressed by the formula

T_{1} = \frac{2\pi (R_{2}-R_{1})}{\mu }\sqrt{\frac{\rho _{0}}{\gamma p_{0}}},

where $\mu$ --- the smallest positive root of the equation

\tg \mu = \frac{\mu }{p\mu ^{2}+1}; \quad p = \frac{R_{1}R_{2}}{(R_{2}-R_{1})^{2}};

here $\gamma$ --- Poisson's constant, $p_{0}$ and $\rho_{0}$ --- the initial pressure and density.

Problem 5864

Prove that the equation of radial vibrations of a gas $\dfrac {\partial^{2} \Phi }{\partial r^{2}}+\dfrac {2}{r}\dfrac {\partial \Phi }{\partial r} = \dfrac {1}{a^{2}}\dfrac {\partial^{2} \Phi }{\partial t^{2}}$ has the solution $\Phi = \dfrac {A}{r}\cos \dfrac {\omega }{a}(at-r)$ , which vanishes as $r\to \infty$ .

Problem 5865

Prove that if the equation of small vibrations of a gas has the form

\frac{\partial ^{2} \Phi }{\partial r^{2}}+\frac{2}{r}\frac{\partial \Phi }{\partial r}+\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial \Phi }{\partial q}\right) = \frac{1}{a^{2}}\frac{\partial ^{2} \Phi }{\partial t^{2}},

then it admits the integral

\Phi = \frac{A\cos \theta }{r}\left(\cos \frac{\omega }{a}(at-r)+\frac{a}{\omega r}\sin \frac{\omega }{a}(at-r)\right),

which vanishes as $r\to \infty$ .

Problem 5866

Prove that the solution of problem 5802 (problem 5802: a perfectly flexible string of length $l$ is suspended by the end $x = l$ , and to the other end $x = 0$ a weight of $P$ kg is attached; the density of the string $\rho = A/\sqrt{l_{1}+x}$ , where $M = 2A\sqrt{l_{1}}$ ; it is required to find the small vibrations of the string about the equilibrium position) is given by the series

u = \sum _{k=1}^{\infty } (a_{k}\cos a\lambda _{k}t + b_{k}\sin a\lambda _{k}t) J_{0}\left(\mu _{k}\sqrt{\frac{x}{l}}\right),

where

a_{k} = \frac{1}{lJ_{1}^{2}(\mu _{k})}\int _{0}^{l} f(x)J_{0}\left(\mu _{k}\sqrt{\frac{x}{l}}\right) \, dx; \quad f(x) = u\big|_{t=0},

b_{k} = \frac{1}{a\mu _{k}\sqrt{l}J_{1}^{2}(\mu _{k})}\int _{0}^{1} F(x)J_{0}\left(\mu _{k}\sqrt{\frac{x}{l}}\right) \, dx, \quad F(x) = \frac{\partial u}{\partial t}\Big|_{t=0},

\lambda _{k} = \sqrt{\frac{\mu _{k}^{2}}{4l}-\left(\frac{\omega }{a}\right)^{2}},

and $\mu_{1}, \mu_{2}, \ldots$ denote the positive roots of the equation $J_{0}(x) = 0$ .

Problem 5867

The linear density of a suspended string varies according to the law $\rho = ax^{m}$ , where $m > -1$ . Derive the equation of small vibrations of the string and prove that the deviation from the equilibrium position is expressed by the formula

u = \sum _{k=1}^{\infty } N_{k}\frac{J_{m}\left(\mu _{k}\sqrt{\frac{x}{l}}\right)}{x^{\frac{m}{2}}}\sin \left(\frac{1}{2}\sqrt{\frac{g}{l(m-1)}}\mu _{k}t+\varphi _{k}\right),

where $\mu_{1}, \mu_{2}, \ldots$ --- the positive roots of the equation $J_{m}(x) = 0$ , and $N_{k}$ and $\varphi_{k}$ --- constants depending on the initial conditions.

Problem 5868

On a circular membrane, fastened at the rim, an external harmonic force acts

\rho Y = \rho A\sin \omega t,

continuously distributed over its entire area.

Prove that the forced vibrations of the membrane are expressed by the equality

u = \frac{A}{\omega ^{2}}\left(\frac{J_{0}\left(\frac{\omega }{a}r\right)}{J_{0}\left(\frac{\omega }{a}R\right)}-1\right)\sin \omega t,

where $R$ --- the radius of the membrane.

Problem 5869

Given a thin homogeneous rod of length $l$ , isolated from the external space. Throughout the entire time of observation the ends $x = 0$ and $x = l$ are held at constant temperature $u = 0$ . In addition, the initial distribution of temperature along the rod is known.

Determine its temperature at the moment of time $t$ (the heat conduction equation of the rod --- problem 5812: $\frac{\partial^{2} u}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial u}{\partial t}$ , $a = \sqrt{k/(c\rho )}$ ).

Remark.

Hint. The problem reduces to integrating the heat conduction equation $\dfrac {\partial u}{\partial t} = a^{2}\dfrac {\partial^{2} u}{\partial x^{2}}$ , $a = \sqrt{k/(c\rho )}$ under the conditions: $u\big|_{x=0} = 0$ , $u\big|_{x=l} = 0$ , for every $t$ ; $u\big|_{t=0} = f(x)$ for $0 < x < l$ .

Problem 5870

Solve the preceding problem (problem 5869) under the assumption that the end $x = 0$ of the rod has temperature equal to zero, and that at the end $x = l$ heat is radiated into the surrounding medium, whose temperature is taken to be zero.

Remark.

Hint. The second of the boundary conditions of the preceding problem should be replaced by the condition: $k\dfrac {\partial u}{\partial x}\Big|_{x=l} = -hu\big|_{x=l}$ , where $k$ and $h$ denote the coefficients of internal and external thermal conductivity of the material of the rod.

Problem 5871

Solve the preceding problem (problem 5870) under the assumption that radiation of heat into the surrounding medium occurs at both ends of the rod.

Remark.

Hint. The boundary conditions have the form

\frac{\partial u}{\partial x}\Big|_{x=0} = \frac{h}{k}u\big|_{x=0}, \quad \frac{\partial u}{\partial x}\Big|_{x=l} = -\frac{h}{k}u\big|_{x=l}.

Problem 5872

Apply the Fourier method to the study of heat propagation in a homogeneous ring, assuming that the initial distribution of heat is known (the heat conduction equation of the ring --- problem 5814: $\frac{1}{R^{2}}\frac{\partial^{2} u}{\partial \theta^{2}}-bu = \frac{1}{a^{2}}\frac{\partial u}{\partial t}$ , $a = \sqrt{k/(c\rho )}$ , $b = hl/(\sigma k)$ ).

Remark.

Hint. The problem reduces to integrating the equation

\frac{\partial v}{\partial t} = a^{2}\frac{\partial ^{2} v}{\partial \theta ^{2}}, \quad u = ve^{-bt}

under the fulfillment of the initial condition

u\big|_{t=0} = f(\theta ) \ \text{for} \ 0 < \theta < 2\pi .

Problem 5873

Given a homogeneous sphere of radius $R$ , whose center is located at the origin of coordinates. It is known that the initial temperature of any point of the sphere depends only on the distance $r$ of that point from the center of the sphere. Throughout the entire time of observation the outer surface of the sphere is held at temperature $0°$ . Determine the temperature of any point inside the sphere at the moment of time $t$ (the equation of radial heat propagation in the sphere --- problem 5816: $\frac{\partial^{2} v}{\partial r^{2}} = \frac{1}{a^{2}}\frac{\partial v}{\partial t}$ , $v(r,t) = ru(r,t)$ , $a = \sqrt{k/(c\rho )}$ ).

Remark.

Hint. The problem reduces to integrating the equation $\dfrac {\partial v}{\partial t} = a^{2}\dfrac {\partial^{2} v}{\partial r^{2}}$ , $u = \dfrac {v}{r}$ , $a = \sqrt{k/(c\rho )}$ and to satisfying the conditions: $v\big|_{r=0} = 0$ , $v\big|_{r=R} = 0$ for every $t$ ; $v\big|_{t=0} = rf(r)$ for $0 < r < R$ .

Problem 5874

Solve the preceding problem (problem 5873) under the assumption that the surface of the sphere is continually freely cooling in a medium having temperature $0°$ .

Remark.

Hint. The second of the boundary conditions in the preceding problem must be changed to the following:

\frac{\partial v}{\partial r}\Big|_{r=R} + \left(H-\frac{1}{R}\right)v\big|_{r=R} = 0, \quad H = \frac{h}{k}.

Problem 5875

A homogeneous gas is enclosed in a fixed vessel having the shape of a sphere of radius $R$ . Investigate the radial vibrations of the gas, assuming that the motion occurs with a velocity potential.

Remark.

Hint. For small vibrations, the velocity potential satisfies the wave equation

\frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}}+\frac{\partial ^{2} u}{\partial z^{2}} = \frac{1}{a^{2}}\frac{\partial ^{2} u}{\partial t^{2}},

$a = \sqrt{\gamma p_{0}/\rho_{0}}$ , where $p_{0}$ and $\rho_{0}$ --- the initial pressure and density, and $\gamma$ --- Poisson's constant.

Transforming the vibration equation to spherical coordinates $r$ , $\theta$ , $\varphi$ , seek by the Fourier method the integral satisfying the conditions $u\big|_{r=0} =$ a finite value,

\frac{\partial u}{\partial r}\Big|_{r=R} = 0 \ \text{for every} \ t, \quad u\big|_{t=0} = f(r), \quad \frac{\partial u}{\partial t}\Big|_{t=0} = F(r)

for $0 < r < R$ .

Problem 5876

A heavy homogeneous string of length $l$ , suspended by one of its ends ( $x = l$ ), is taken out of equilibrium and released without initial velocity (Fig. 15.17). Determine the deviation $u$ as a function of the abscissa $x$ and the time $t$ .

Fig. 15.17

Remark.

Hint. The problem reduces to integrating the equation $x\dfrac {\partial^{2} u}{\partial x^{2}}+\dfrac {\partial u}{\partial x} = \dfrac {1}{a^{2}}\dfrac {\partial^{2} u}{\partial t^{2}}$ , $a = \sqrt{g}$ under the conditions: $u\big|_{x=0} =$ a finite value, $u\big|_{x=l} = 0$ for every $t$ , $u\big|_{t=0} = f(x)$ , $\dfrac {\partial u}{\partial t}\Big|_{t=0} = 0$ for $0 < x < l$ .

Problem 5877

Given a cylindrical tube of radius $R$ , long enough that it can be considered as extending in both directions to infinity. Investigate the small radial vibrations of a homogeneous gas enclosed in the tube.

Remark.

Hint. The vibration equation should be transformed to cylindrical coordinates $r$ , $\varphi$ , $z$ , directing the axis $z$ along the axis of the tube, and then one should seek the integral of the equation

\frac{\partial ^{2} u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r} = \frac{1}{a^{2}}\frac{\partial ^{2} u}{\partial t^{2}},

satisfying the conditions: $u\big|_{r=0} =$ a finite value, $\dfrac {\partial u}{\partial r}\Big|_{r=R} = 0$ for every $t$ ; $u\big|_{t=0} = f(r)$ , $\dfrac {\partial u}{\partial t}\Big|_{t=0} = F(r)$ for $0 < r < R$ .

Problem 5878

Apply the Fourier method to the study of the radial vibrations of a membrane, having the shape of a circle of radius $R$ and fastened at the rim (the equation of small vibrations of the membrane --- problem 5811: $\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}} = \frac{1}{a^{2}}\frac{\partial^{2} u}{\partial t^{2}}$ , $a = \sqrt{T_{0}/\sigma }$ ).

Remark.

Hint. The problem reduces to integrating the equation

\frac{\partial ^{2} u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r} = \frac{1}{a^{2}}\frac{\partial ^{2} u}{\partial t^{2}},

$a = \sqrt{T_{0}/\sigma }$ under the conditions $u\big|_{r=0} =$ a finite value, $u\big|_{r=R} = 0$ for every $t$ and $u\big|_{t=0} = f(r)$ , $\dfrac {\partial u}{\partial t}\Big|_{t=0} = F(r)$ for $0 < r < R$ .

Problem 5879

Investigate the radial propagation of heat in an infinite cylinder of radius $R$ , whose lateral surface is at all times at temperature $0°$ .

Remark.

Hint. The heat conduction equation in cylindrical coordinates has the form

\frac{\partial u}{\partial t} = a^{2}\left(\frac{\partial ^{2} u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}\right), \quad a = \sqrt{\frac{k}{c\rho }}.

One should seek the integral of this equation satisfying the conditions $u\big|_{r=0} =$ a finite value, $u\big|_{r=R} = 0$ for every $t$ and $u\big|_{t=0} = f(r)$ for $0 < r < R$ .

Problem 5880

A cylinder with base radius $R$ and height $h$ has, throughout the entire experiment, the temperature of its lower base and lateral surface equal to $0°$ , while the temperature of the upper base is a certain function of $r$ . Find the steady-state temperature of the interior points of the cylinder.

Remark.

Hint. To solve the problem it is necessary to find such an integral of the equation $\dfrac {\partial^{2} u}{\partial r^{2}}+\dfrac {1}{r}\dfrac {\partial u}{\partial r}+\dfrac {\partial^{2} u}{\partial z^{2}} = 0$ as satisfies the conditions: $u\big|_{r=0} =$ a finite value, $u\big|_{z=0} = 0$ , $u\big|_{r=R} = 0$ , $u\big|_{z=h} = f(r)$ for $0 < r < R$ .

Problem 5881

Solve the preceding problem (problem 5880) under the assumption that the lateral surface of the cylinder is covered with a jacket impermeable to heat.

Remark.

Hint. The third of the boundary conditions given in the preceding problem should be replaced by the following: $\dfrac {\partial u}{\partial r}\Big|_{r=R} = 0$ .

Problem 5882

Solve problem 5880 under the assumption that the lateral surface of the cylinder is freely cooling in air having temperature $0°$ .

Remark.

Hint. The boundary condition on the lateral surface has the form: $\dfrac {\partial u}{\partial r}\Big|_{r=R}+h_{1}u\big|_{r=R} = 0$ , where $h_{1} = \dfrac {h}{k}$ ; for the meanings of the letters $h$ and $k$ see problem 5870 (problem 5870: $k$ and $h$ --- coefficients of internal and external thermal conductivity of the material of the rod).

Problem 5883

A cylinder with base radius $R$ and height $h$ has the temperature of both bases equal to $0°$ , while the temperature of the lateral surface represents a given function of $z$ : $u = f(z)$ . Find the steady-state temperature of the interior points of the cylinder.

Problem 5884

Given a thin rectangular plate with sides $l$ and $m$ (coinciding with the coordinate axes), for which the initial distribution of temperature is known. Throughout the entire time of observation the perimeter of the plate is held at temperature $0°$ . It is required to determine the temperature of each point of the plate after a time $t$ from the initial moment (the heat conduction equation of the plate --- problem 5813: $\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}} = \frac{1}{a^{2}}\frac{\partial u}{\partial t}$ , $a = \sqrt{k/(c\rho )}$ ).

Remark.

Hint. The problem reduces to integrating the equation

\frac{\partial u}{\partial t} = a^{2}\left(\frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}}\right), \quad a = \sqrt{\frac{k}{c\rho }}

under the conditions:

u = 0 \ \text{at} \ x = 0, \quad u = 0 \ \text{at} \ x = l \quad \text{for} \ 0 < y < m \ \text{and for every} \ t,

u = 0 \ \text{at} \ y = 0, \quad u = 0 \ \text{at} \ y = m \quad \text{for} \ 0 < x < l \ \text{and for every} \ t,

u\big|_{t=0} = f(x,y) \quad \text{for} \ 0 < x < l \ \text{and} \ 0 < y < m.

Problem 5885

Given a rectangular parallelepiped with edges $l$ , $m$ , $n$ (three of its edges coinciding with the coordinate axes). Throughout the entire time of observation the lateral faces are held at temperature $0°$ , while both bases have a given distribution of temperature. Determine the steady-state temperature of the interior points of the parallelepiped.

Remark.

Hint. The desired function satisfies the equation $\dfrac {\partial^{2} u}{\partial x^{2}}+\dfrac {\partial^{2} u}{\partial y^{2}}+\dfrac {\partial^{2} u}{\partial z^{2}} = 0$ under the conditions: $u = 0$ at $x = 0$ , $u = 0$ at $x = l$ , $u = 0$ at $y = 0$ , $u = 0$ at $y = m$ , $u = f(x,y)$ at $z = 0$ , $u = F(x,y)$ at $z = n$ .

Problem 5886

Investigate the small vibrations of a membrane, fastened at the edges and having the shape of a rectangle, whose sides $l$ and $m$ coincide with the coordinate axes $x$ and $y$ (the equation of small vibrations of the membrane --- problem 5811: $\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}} = \frac{1}{a^{2}}\frac{\partial^{2} u}{\partial t^{2}}$ , $a = \sqrt{T_{0}/\sigma }$ ).

Remark.

Hint. The problem reduces to integrating the equation

\frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}} = \frac{1}{a^{2}}\frac{\partial ^{2} u}{\partial t^{2}}, \quad a = \sqrt{\frac{T_{0}}{\sigma }}

under the conditions:

u = 0 \ \text{at} \ x = 0, \qquad u = 0 \ \text{at} \ x = l,

u = 0 \ \text{at} \ y = 0, \qquad u = 0 \ \text{at} \ y = m.

u\big|_{t=0} = f(x,y), \quad \frac{\partial u}{\partial t}\Big|_{t=0} = F(x,y) \quad \text{for} \ 0 < x < l, \ 0 < y < m.

Problem 5887

Investigate the propagation of heat in an infinitely long rod, isolated from the external space (the heat conduction equation of the rod --- problem 5812: $\frac{\partial^{2} u}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial u}{\partial t}$ , $a = \sqrt{k/(c\rho )}$ ).

Remark.

Hint. The desired temperature satisfies the equation

\frac{\partial u}{\partial t} = a^{2}\frac{\partial ^{2} u}{\partial x^{2}}, \quad a = \sqrt{\frac{k}{c\rho }}

and the condition $u\big|_{t=0} = f(x)$ .

Problem 5888

Given a homogeneous rod, one end of which is removed to infinity, while the other has a temperature that changes with time in a known manner. Determine the temperature of the rod after a time $t$ from the initial moment.

Remark.

Hint. One should integrate the equation of heat propagation in the rod under the conditions: $u\big|_{x=0} = \varphi (t)$ , $u\big|_{t=0} = f(x)$ .

Problem 5889

Investigate the propagation of heat in a thin plate, assuming that it is so large that it can be considered as extending unboundedly in all directions (the heat conduction equation of the plate --- problem 5813: $\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}} = \frac{1}{a^{2}}\frac{\partial u}{\partial t}$ , $a = \sqrt{k/(c\rho )}$ ).

Remark.

Hint. The desired temperature satisfies the equation

\frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}} = \frac{1}{a^{2}}\frac{\partial u}{\partial t}, \quad a = \sqrt{\frac{k}{c\rho }}

and the condition $u\big|_{t=0} = f(x,y)$ .

Problem 5890

Investigate the propagation of heat in an infinite plate, bounded on one edge, on which the temperature changes in a known manner.

Remark.

Hint. If the axis of ordinates is drawn along the edge of the plate, then the problem reduces to integrating the equation given in the preceding problem, subject to the conditions:

u\big|_{x=0} = \varphi (y,t); \quad u\big|_{t=0} = f(x,y).

Problem 5891

Determine the temperature at the point $M(x,y,z)$ of unbounded space, assuming that the initial distribution of temperature is known.

Remark.

Hint. The desired temperature satisfies the equation

\frac{\partial u}{\partial t} = a^{2}\left(\frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}}+\frac{\partial ^{2} u}{\partial z^{2}}\right), \quad a = \sqrt{\frac{k}{c\rho }}

and the condition $u\big|_{t=0} = f(x,y,z)$ .

Problem 5892

A flexible homogeneous string rotates with constant angular velocity $\omega$ about a vertical axis. At the moment of time taken as the initial one, the string is taken out of its relative equilibrium position, and its points are imparted initial velocities. Determine the deviation $u(x,t)$ of the string after a time $t$ from the initial moment (the equation of vibrations of the string --- problem 5804: $\frac{\partial }{\partial x}\left((l^{2}-x^{2})\frac{\partial u}{\partial x}\right) = \frac{1}{a^{2}}\frac{\partial^{2} u}{\partial t^{2}}$ , $a = \omega /\sqrt{2}$ ).

Remark.

Hint. The problem reduces to integrating the equation

(l^{2}-x^{2})\frac{\partial ^{2} u}{\partial x^{2}} - 2x\frac{\partial u}{\partial x} = \frac{1}{a^{2}}\frac{\partial ^{2} u}{\partial t^{2}}, \quad a = \frac{\omega }{\sqrt{2}}

under the conditions: $u\big|_{x=0} = 0$ , $u\big|_{x=l} =$ a finite value for every $t$ and $u\big|_{t=0} = f(x)$ , $\dfrac {du}{dt}\Big|_{t=0} = F(x)$ for $0 < x < l$ .

Problem 5893

Prove that Laplace's equation

\frac{\partial ^{2} u}{\partial r^{2}}+\frac{2}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial ^{2} u}{\partial \theta ^{2}}+\frac{\ctg \theta }{r^{2}}\frac{\partial u}{\partial \theta } = 0

admits the particular solution

u = \left(Ar^{m}+\frac{B}{r^{m+1}}\right)P_{m}(\cos \theta ),

where $A$ and $B$ denote arbitrary constants, $m$ is a nonnegative integer, and $P_{m}(\cos \theta )$ is the Legendre polynomial of index $m$ .

Problem 5894

A given function $f(r)$ can be expanded into series of the following form:

f(r) = A_{0}+A_{1}r+A_{2}r^{2}+\ldots +A_{n}r^{n}+\ldots \quad \text{for } r < R,

f(r) = \frac{B_{1}}{r}+\frac{B_{2}}{r^{2}}+\frac{B_{3}}{r^{3}}+\ldots +\frac{B_{n+1}}{r^{n+1}}+\ldots \quad \text{for } r > R.

Find such a solution of the equation given in the preceding problem (problem 5893: $\frac{\partial^{2} u}{\partial r^{2}}+\frac{2}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2} u}{\partial \theta^{2}}+\frac{\ctg \theta }{r^{2}}\frac{\partial u}{\partial \theta } = 0$ , with particular solution $u = \left(Ar^{m}+\frac{B}{r^{m+1}}\right)P_{m}(\cos \theta )$ ) as would satisfy the condition

u\big|_{\theta =0} = f(r).

Problem 5895

A circular wire ring of radius $R$ is charged with $E$ units of static electricity. Find the potential $v(r,\theta )$ of the electrified ring at any point of the field (Fig. 15.18).

Fig. 15.18

Remark.

Hint. Use the result of the preceding problem.

Problem 5896

Find such a solution of the equation of problem 5893 (problem 5893: $\frac{\partial^{2} u}{\partial r^{2}}+\frac{2}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2} u}{\partial \theta^{2}}+\frac{\ctg \theta }{r^{2}}\frac{\partial u}{\partial \theta } = 0$ , with particular solution $u = \left(Ar^{m}+\frac{B}{r^{m+1}}\right)P_{m}(\cos \theta )$ ) that, for $r = R$ (on the surface of a sphere of radius $R$ centered at the origin), it turns into a given function of the angle $\theta$ , i.e. $u\big|_{r=R} = f(\theta )$ for $0 \leqslant \theta \leqslant \pi$ .

Remark.

Hint. Take into account the result of problem 5893 and use the expansion of the function $f(\theta )$ in a series of Legendre polynomials

f(\theta ) = \sum _{n=0}^{\infty } A_{n}P_{n}(\cos \theta ),

A_{n} = \frac{2n+1}{2}\int _{0}^{\pi } f(\theta )P_{n}(\cos \theta )\sin \theta \, d\theta .

Problem 5897

By a layer of non-conducting substance, the surface of a spherical conductor of radius $R$ is divided into two hemispheres, charged to potentials: $v_{1}$ (upper) and $v_{2}$ (lower). Find the potential $v(r,\theta )$ of such a conductor at any point of the field (Fig. 15.19).

Fig. 15.19

Remark.

Hint. One should use the result of the preceding problem, taking into account the equalities:

v\big|_{r=R} = f(\theta ) = \begin{cases} v_{1} & \text{for } 0 \leqslant \theta \leqslant \dfrac {\pi }{2}, \\ v_{2} & \text{for } \dfrac {\pi }{2} \leqslant \theta \leqslant \pi . \end{cases}

Problem 5898

A metal ball of radius $R$ with a soot-blackened surface is subject to the action of sunlight in air whose temperature equals $0°$ . Determine the steady-state temperature of the interior points of the ball (Fig. 15.20).

Fig. 15.20

Remark.

Hint. The problem reduces to integrating Laplace's equation (see problem 5893: $\frac{\partial^{2} u}{\partial r^{2}}+\frac{2}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2} u}{\partial \theta^{2}}+\frac{\ctg \theta }{r^{2}}\frac{\partial u}{\partial \theta } = 0$ ) under the following condition:

\frac{\partial u}{\partial r}\Big|_{r=R} + H\left(u-f(\theta )\right)\Big|_{r=R} = 0,

where $H$ is constant, and

f(\theta ) = \begin{cases} M\cos \theta , & \text{for } 0 \leqslant \theta \leqslant \dfrac {\pi }{2} \quad (M = \text{const}); \\ 0, & \text{for } \dfrac {\pi }{2} \leqslant \theta \leqslant \pi . \end{cases}

Problem 5899

Integrate by the Fourier method the equation of transverse vibrations of a beam $\dfrac {\partial^{4} u}{\partial x^{4}}+\dfrac {1}{a^{2}}\dfrac {\partial^{2} u}{\partial t^{2}} = 0$ under the conditions $u = 0$ , $\dfrac {\partial u}{\partial x} = 0$ at $x = 0$ , $x = l$ , expressing that both ends of the beam are rigidly fastened (Fig. 15.21).

Fig. 15.21

Remark.

Hint. The characteristic numbers $\lambda_{1}, \lambda_{2}, \lambda_{3}, \ldots$ of the problem are determined from the equation

\ch \lambda l \cdot \cos \lambda l = 1,

and the corresponding fundamental functions

X_{n}(x) = (\sh \lambda _{n}l - \sin \lambda _{n}l)(\ch \lambda _{n}x - \cos \lambda _{n}x) - (\ch \lambda _{n}l - \cos \lambda _{n}l)(\sh \lambda _{n}x - \sin \lambda _{n}x)

form a system of functions orthogonal on the interval $(0,l)$ .

Section 15.6Integral Equations[61/80%]

Problem 5900

At the point $x = \xi$ of a homogeneous string ( $\rho = 1$ ), subject to the action of a strong tension $T_{0}$ and fastened at the ends $x = 0$ and $x = 1$ , a weight $Q = 1$ is suspended (Fig. 15.22).

Fig. 15.22

Prove that the equation of the sag of the string has the form

u = K(x,\xi ) = \begin{cases} x(1-\xi ) \cdot \dfrac {1}{T_{0}} & \text{for } x \leqslant \xi , \\ \xi (1-x) \cdot \dfrac {1}{T_{0}} & \text{for } x \geqslant \xi . \end{cases}

Problem 5901

A heavy homogeneous string is fastened at the point $x = 1$ and is subject at the point $x = \xi$ to the action of a horizontal force $Q = 1$ (Fig. 15.23).

Fig. 15.23

Prove that the equilibrium equation of the string has the form

u = K(x,\xi ) = \begin{cases} -\ln \xi , & \text{for } x \leqslant \xi , \\ -\ln x, & \text{for } x \geqslant \xi . \end{cases}

Remark.

Hint. The tension of the string at the point $M(x)$ is taken equal to $-x$ .

Problem 5902

A homogeneous string of length $l = 1$ is fastened by the end $x = 0$ to a fixed axis and rotates about this axis with constant angular velocity. To the point $x = \xi$ of such a string a force $Q = 1$ is applied (Fig. 15.24).

Fig. 15.24

Prove that the equation of relative equilibrium of the string has the form

u = K(x,\xi ) = \begin{cases} \dfrac {1}{2}\ln \dfrac {1+x}{1-x} & \text{for } x \leqslant \xi , \\ \dfrac {1}{2}\ln \dfrac {1+\xi }{1-\xi } & \text{for } x \geqslant \xi . \end{cases}

Remark.

Hint. The tension of the string at the point $M(x)$ , caused by the centrifugal force, is taken equal to

R = 2\int _{x}^{1} x \, dx = 1-x^{2}.

Problem 5903

Prove that for a continuously distributed load, the equation of the sag of the string has the form

u = \int _{0}^{1} K(x,\xi )p(\xi ) \, d\xi ,

where $p(\xi )$ denotes the magnitude of the load, computed per unit length of the string.

Remark.

Hint. Use the result of problem 5900 (problem 5900: $K(x,\xi ) = x(1-\xi )/T_{0}$ for $x \leqslant \xi$ , $K(x,\xi ) = \xi (1-x)/T_{0}$ for $x \geqslant \xi$ ).

Problem 5904

Prove that the problem of small vibrations of a fastened string can be reduced to solving the homogeneous Fredholm equation

y(x) = \lambda \int _{0}^{1} K(x,\xi )y(\xi ) \, d\xi ,

where the symmetric kernel $K(x,\xi )$ has the form given in problem 5901 (problem 5901: $K(x,\xi ) = -\ln \xi$ for $x \leqslant \xi$ , $K(x,\xi ) = -\ln x$ for $x \geqslant \xi$ ).

Problem 5905

On a homogeneous string, fastened at the ends, a continuously distributed force acts

F(\xi ,t) = p_{0}(\xi )\sin \omega t.

Prove that the problem of forced vibrations of such a string reduces to solving the inhomogeneous Fredholm equation

y(x) = f(x) + \lambda \int _{0}^{1} K(x,\xi )y(\xi ) \, d\xi .

Problem 5906

Considering the sequence $1, x, x^{2}, \ldots , x^{n}, \ldots$ on the interval $(-1,+1)$ , replace it by an orthogonal and normalized sequence, taking as the weight function

p(x) = 1.

Write out the first three terms of the sequence.

Remark.

Remark. Considering the quadratic error

\int _{a}^{b} p(x)\left(f(x)-s_{n}(x)\right)^{2} \, dx,

we call $p(x)$ the weight.

Problem 5907

Solve problem 5905 (problem 5905: on a homogeneous string, fastened at the ends, a continuously distributed force $F(\xi ,t) = p_{0}(\xi )\sin \omega t$ acts; the problem of forced vibrations reduces to solving the inhomogeneous Fredholm equation $y(x) = f(x)+\lambda \int_{0}^{1} K(x,\xi )y(\xi )\, d\xi$ ), taking as the weight function

p(x) = \frac{1}{\sqrt{1-x^{2}}}.

Problem 5908

Considering the sequence $1, x, x^{2}, \ldots , x^{n}, \ldots$ on the interval $(0,\infty )$ , replace it by an orthogonal and normalized one, taking the weight function to be

p(x) = e^{-x}.

Write out the first three terms of the sequence.

Problem 5909

Considering the sequence $1, x, x^{2}, \ldots , x^{n}, \ldots$ on the interval $(-\infty ,+\infty )$ , replace it by an orthogonal and normalized one, taking the weight function to be

p(x) = e^{-x^{2}}.

Write out the first three terms of the sequence.

Remark.

Hint. Use the properties of the function $\Gamma (p)$ .

Problem 5910

Reduce to an integral equation the problem of finding the solutions of the equation

y''+xy'+y = 0, \quad y(0) = 1, \quad y'(0) = 0.

Problem 5911

Find $u(x)$ from the equation

u(x) = x+4\int _{0}^{x} (t-x)u(t) \, dt.

Problem 5912

Replace the finding of $y$ from the equation $y''+y = 0$ with the initial condition $y(0) = 0$ , $y'(0) = -1$ by solving an integral equation.

Problem 5913

Replace the finding of $y$ from the equation $y''-5y'+6y = 0$ with the initial conditions $y(0) = 0$ , $y'(0) = 1$ by solving an integral equation.

Problem 5914

Replace the finding of $y$ from the equation $y''+y = \cos x$ with the initial conditions $y(0) = y'(0) = 0$ by solving an integral equation.

Problem 5915

Solve the equation

\varphi (x) = \frac{1}{2}\int _{0}^{1} xt\varphi (t) \, dt + \frac{5x}{6}.

Problem 5916

Solve the equation

\varphi (x) = 3\int _{0}^{1} xt\varphi (t) \, dt + 3x-2.

Problem 5917

Solve the equation

\varphi (x) = \frac{1}{2}\int _{0}^{2} xt\varphi (t) \, dt + \frac{5x}{6}.

Problem 5918

Solve the equation

\varphi (x) = 3\int _{0}^{2} xt\varphi (t) \, dt + 3x-2.

Problem 5919

Find the Fredholm determinant and the resolvent for the equation

\varphi (x) = \lambda \int _{a}^{b} K(x,y)\varphi (y) \, dy + F(x),

as well as the first characteristic number and its fundamental function, if

K(x,y) = g(x)g(y);

solve the equation, assuming that $\lambda$ is not a characteristic number.

Problem 5920

The same (find the Fredholm determinant and the resolvent for the equation $\varphi (x) = \lambda \int_{a}^{b} K(x,y)\varphi (y)\, dy+F(x)$ , as well as the first characteristic number and its fundamental function; solve the equation, assuming that $\lambda$ is not a characteristic number), assuming

K(x,y) = g(y).

Problem 5921

K(x,y) = g(x).

Problem 5922

Solve the equation

\varphi (x) = \frac{2}{e^{2}-1}\int _{0}^{1} e^{x}e^{y}\varphi (y) \, dy.

Problem 5923

Solve the equation

\varphi (x) = x^{2}+2\int _{0}^{1} (1+3xy)\varphi (y) \, dy.

Problem 5924

Find all solutions of the equation

\varphi (x) = \lambda \int _{0}^{1} y\varphi (y) \, dy + F(x).

Problem 5925

In the equation

\varphi (x) = \lambda \int _{0}^{1} K(x,y)\varphi (y) \, dy + F(x)

K(x,y) = \begin{cases} x, & \text{if } x < y; \\ y, & \text{if } x > y. \end{cases}

Find the Fredholm determinant and the resolvent.

Problem 5926

Find the Fredholm determinant and the resolvent for the equation

\varphi (x) = \lambda \int _{a}^{b} K(x,y)\varphi (y) \, dy + F(x),

as well as the first characteristic number and its fundamental function, if

K(x,y) = c.

Problem 5927

For the equation

\varphi (x) = \lambda \int _{0}^{1} (x+y)\varphi (y) \, dy + F(x)

find the characteristic numbers and fundamental functions.

Problem 5928

Solve the equation

\varphi (x) = \int _{0}^{1} (x+y)\varphi (y) \, dy + 18x^{2}-9x-4.

Remark.

Hint. See problem 5927 (problem 5927: for the equation $\varphi (x) = \lambda \int_{0}^{1} (x+y)\varphi (y)\, dy+F(x)$ find the characteristic numbers and fundamental functions).

Problem 5929

For the equation

\varphi (x) = \lambda \int _{0}^{2\pi } (\sin x\sin y + \sin 2x\sin 2y)\varphi (y) \, dy + F(x)

write out the Fredholm determinant, its minors, the resolvent, and the fundamental functions.

Problem 5930

For the equation

\varphi (x) = \lambda \int _{0}^{\pi } \cos (x+y)\varphi (y) \, dy + F(x)

write out the Fredholm determinant, its minor, the resolvent, the fundamental functions, and the solution.

Problem 5931

Solve the equation

\varphi (x) = \int _{0}^{\pi } \cos (x+y)\varphi (y) \, dy + 1.

Remark.

Hint. See problem 5930 (problem 5930: for the equation $\varphi (x) = \lambda \int_{0}^{\pi } \cos (x+y)\varphi (y)\, dy+F(x)$ write out the Fredholm determinant, its minor, the resolvent, the fundamental functions, and the solution).

Problem 5932

Find the resolvent of the equation

\varphi (x) = \lambda \int _{0}^{1} \left(xy-\frac{x+y}{2}+\frac{1}{3}\right)\varphi (y) \, dy;

find the fundamental functions.

Problem 5933

Solve the equation

\varphi (x) = 6\int _{0}^{1} \left(xy-\frac{x+y}{2}+\frac{1}{3}\right)\varphi (y) \, dy + x.

Problem 5934

Find the resolvent of the equation

\varphi (x) = \lambda \int _{0}^{1} (x^{2}+y^{2})\varphi (y) \, dy + F(x).

Problem 5935

Find the resolvent of the equation

\varphi (x) = \lambda \int _{0}^{1} (x^{2}y+y^{2}x)\varphi (y) \, dy + F(x).

Problem 5936

Find the resolvent of the equation

\varphi (x) = \lambda \int _{0}^{1} (1+x+y)\varphi (y) \, dy + F(x).

Problem 5937

Find the resolvent of the equation

\varphi (x) = \lambda \int _{0}^{1} (x-y)\varphi (y) \, dy + F(x).

Problem 5938

Verify that the equation

\varphi (x) = \lambda \int _{-\infty }^{+\infty } e^{-\left|x-y\right|}\varphi (y) \, dy

has a solution for every $\lambda > \dfrac {1}{2}$ .

Remark.

Hint. Form a differential equation for $\varphi (x)$ .

Problem 5939

Solve the system of equations

\varphi (x) = \lambda \int _{0}^{1} (x-y)\psi (y) \, dy, \quad \psi (x) = \lambda \int _{0}^{1} (y-x)\varphi (y) \, dy.

Remark.

Hint. See problem 5932 (problem 5932: find the resolvent of the equation $\varphi (x) = \lambda \int_{0}^{1} \left(xy-\frac{x+y}{2}+\frac{1}{3}\right)\varphi (y)\, dy$ and its fundamental functions).

Problem 5940

Given that

K(x,y) = \begin{cases} -\dfrac {1}{2}, & \text{if } x < y, \\ \dfrac {1}{2}, & \text{if } x > y. \end{cases}

Solve the system of equations

\varphi (x) = \pi \int _{0}^{1} K(x,y)\psi (y) \, dy,

\psi (x) = \pi \int _{0}^{1} K(y,x)\varphi (y) \, dy.

Problem 5941

Reduce to an integral equation the problem of finding the solutions of the equation

(xy')'+\lambda xy = 0

under the boundary conditions: $y(0)$ bounded, $y(1) = 0$ .

Problem 5942

Reduce to an integral equation the problem of finding the solutions of the equation

(xy')'+\left(\lambda x-\frac{n^{2}}{x}\right)y = 0

under the boundary conditions: $y(0) = 0$ , $y(1) = 0$ .

Problem 5943

Construct the integral equation whose fundamental functions are the Legendre functions of order $m$ :

\left((1-x^{2})y'\right)'+\left(n(n+1)-\frac{m^{2}}{1-x^{2}}\right)y = 0;

boundary condition: $y$ bounded at $x = -1$ and $x = +1$ .

Problem 5944

Solve problem 5943 for the case $m = 0$ of the Legendre polynomials.

Remark.

Hint. Consider the equation

\left((1-x^{2})y'\right)'+n(n+1)y = \frac{1}{2}.

Problem 5945

Construct the integral equation whose fundamental functions are the Hermite polynomials:

y''+(1-x^{2})y+\lambda y = 0,

$y = 0$ at $x = \pm \infty$ .

Remark.

Hint. Set $\lambda = -2+\mu$ .

Problem 5946

Construct the integral equation whose fundamental functions are the Laguerre polynomials:

xy''+(1-x)y'+\lambda y = 0;

$y$ bounded at $x = 0$ and $x = +\infty$ .

Remark.

Hint. Set $\lambda = -1+\mu$ .

Problem 5947

Having found the corresponding Green's function, replace the equation

y''+k^{2}y = f(x), \quad y(0) = y(1) = 0

by an integral one.

Problem 5948

Having found the corresponding Green's function, replace the equation

y''+k^{2}y = f(x), \quad y'(0) = hy(0), \quad y'(1) = -Hy(1)

by an integral equation.

Problem 5949

Find the characteristic numbers and fundamental functions of the equation

\varphi (x) = -\frac{\lambda }{2}\int _{0}^{1} \varphi (y)\left|x-y\right| \, dy + F(x).

Remark.

Hint. Having determined the boundary conditions, reduce the finding of the characteristic numbers to integrating an ordinary equation.

Problem 5950

Having found the corresponding Green's function $G(\xi ,x)$ , replace the equation

y''+k^{2}y = f(x), \quad y(0) = 0, \quad y'(1) = -hy(1)

by an integral one.

Problem 5951

Having verified that the fundamental functions of the equation of problem 5950 (problem 5950: $y''+k^{2}y = f(x)$ , $y(0) = 0$ , $y'(1) = -hy(1)$ , with Green's function $G(\xi ,x)$ ) equal

\sqrt{\frac{2(h^{2}+k_{n}^{2})}{h^{2}+h+k_{n}^{2}}}\sin k_{n}x,

where $k_{n}$ --- the root of the equation

k = -h\tg k,

solve the equation

y(x) = k^{2}\int _{0}^{1} G(\zeta ,x)y(\zeta ) \, d\zeta + 1.

Problem 5952

Verify that for the equation

\frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}} = 0,

in the case of a disk of radius $R$ with boundary condition $u = 0$ , Green's function equals

G(x,y,\zeta ,\eta ) = -\frac{1}{2\pi }\ln \frac{r_{1}}{r_{2}} + \frac{1}{2\pi }\ln \frac{\sqrt{\zeta ^{2}+\eta ^{2}}}{R},

where $r_{1}$ and $r_{2}$ --- the distances from the point $(x,y)$ to the points

(\zeta ,\eta ) \quad \text{and} \quad \left(\frac{\zeta R^{2}}{\zeta ^{2}+\eta ^{2}}, \frac{\eta R^{2}}{\zeta ^{2}+\eta ^{2}}\right).

Verify the symmetry of the function $G(x,y,\zeta ,\eta )$ .

Problem 5953

Verify that for the equation

\frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}}+\frac{\partial ^{2} u}{\partial z^{2}} = 0

in the case of a sphere of radius $R$ with boundary condition $u = 0$ , Green's function equals

G(x,y,z,\xi ,\eta ,\zeta ) = \frac{1}{4\pi }\left(\frac{1}{r_{1}}-\frac{R}{r_{2}\sqrt{\xi ^{2}+\eta ^{2}+\zeta ^{2}}}\right),

where $r_{1}$ and $r_{2}$ --- the distances from the point $(x,y,z)$ to the points $(\xi ,\eta ,\zeta )$ and

\left(\frac{\xi R^{2}}{\xi ^{2}+\eta ^{2}+\zeta ^{2}}, \frac{\eta R^{2}}{\xi ^{2}+\eta ^{2}+\zeta ^{2}}, \frac{\zeta R^{2}}{\xi ^{2}+\eta ^{2}+\zeta ^{2}}\right).

Problem 5954

Verify that the characteristic numbers and fundamental functions of the equation

\varphi (x,y) = \lambda \iint _{\Omega } G(x,y,\zeta ,\eta )\varphi (\zeta ,\eta ) \, d\zeta \, d\eta ,

where $G$ --- Green's function of problem 5952 (problem 5952: $G(x,y,\zeta ,\eta ) = -\frac{1}{2\pi }\ln \frac{r_{1}}{r_{2}}+\frac{1}{2\pi }\ln \frac{\sqrt{\zeta^{2}+\eta^{2}}}{R}$ for a disk of radius $R$ with boundary condition $u=0$ ), and $\Omega$ --- a disk of radius $R$ , can be found by finding the characteristic numbers and fundamental functions of the equation

\psi (l) = -\lambda \int _{0}^{R} \rho J(\rho ,l)\psi (\rho ) \, d\rho ,

where

J(\rho ,l) = \begin{cases} \ln l - \ln R, & \rho < l; \\ \ln \rho - \ln R, & \rho > l, \end{cases}

$\rho$ and $l$ --- the distances from the points $(\zeta ,\eta )$ , $(x,y)$ to the center of the disk.

Remark.

Hint. Pay attention to the symmetry of the boundary conditions relative to the center of the disk.

Problem 5955

Find the characteristic numbers and fundamental functions of the equation of problem 5954 (problem 5954: $\varphi (x,y) = \lambda \iint_{\Omega } G(x,y,\zeta ,\eta )\varphi (\zeta ,\eta )\, d\zeta \, d\eta$ , where $G$ --- Green's function of problem 5952 for a disk of radius $R$ ).

Problem 5956

Find the characteristic numbers and fundamental functions of the equation

V(x,y,z) = -\frac{\lambda }{4\pi } \iiint _{\Omega } V(\xi ,\eta ,\zeta )G(x,y,z,\xi ,\eta ,\zeta ) \, d\xi \, d\eta \, d\zeta + F(x,y,z),

where $G(x,y,z,\xi ,\eta ,\zeta )$ --- Green's function of problem 5953 (problem 5953: $G(x,y,z,\xi ,\eta ,\zeta ) = \frac{1}{4\pi }\left(\frac{1}{r_{1}}-\frac{R}{r_{2}\sqrt{\xi^{2}+\eta^{2}+\zeta^{2}}}\right)$ for a sphere of radius $R$ with boundary condition $u=0$ ), and $\Omega$ --- a sphere of radius $R$ .

Remark.

Hint. Using the symmetry relative to the center, one can conclude that the fundamental functions depend only on the distance to the origin of coordinates. Use the second-order differential equation defining these functions.

Problem 5957

Find the solution of the equation

\Delta V = \lambda V + F(a),

where $a = \sqrt{x^{2}+y^{2}+z^{2}}$ , satisfying on the boundary of a sphere of radius $R$ centered at the origin the condition $V = 0$ ; the function $V$ must be given inside the sphere.

Remark.

Hint. See problem 5956 (problem 5956: find the characteristic numbers and fundamental functions of the equation $V(x,y,z) = -\frac{\lambda }{4\pi }\iiint_{\Omega } V(\xi ,\eta ,\zeta )G(x,y,z,\xi ,\eta ,\zeta )\, d\xi \, d\eta \, d\zeta +F(x,y,z)$ , where $\Omega$ --- a sphere of radius $R$ ).

Problem 5958

The fundamental functions of the equation

V(x,y,z) = \frac{\lambda }{4\pi } \iiint _{\Omega } \frac{V(\xi ,\eta ,\zeta )}{r} \, d\xi \, d\eta \, d\zeta + F(x,y,z),

where $r$ --- the distance between the points $(x,y,z)$ and $(\xi ,\eta ,\zeta )$ , are called the universal Korn functions.

Find the universal Korn functions, assuming that the region $\Omega$ is a sphere of radius $R$ .

Remark.

Hint. Owing to symmetry, the desired functions depend only on the distance $a$ from the point to the origin of coordinates ( $a = \sqrt{x^{2}+y^{2}+z^{2}}$ ). The function is expressed linearly through the solutions of a certain differential equation, and the undetermined coefficients, together with the characteristic numbers, are found by substitution into the given equation. Normalization is achieved by using the differential equation.

Problem 5959

Prove that the integral equation

u(x) = f(x) + \lambda \int _{0}^{\infty } J_{\nu }(xt)tu(t) \, dt, \quad \nu > -\frac{1}{2}

has the following solution:

u(x) = \frac{f(x)+\lambda \int _{0}^{\infty } J_{\nu }(xt)tf(t) \, dt}{1-\lambda ^{2}}.

Remark.

Hint. One should use the Fourier--Hankel formula:

f(x) = \int _{0}^{\infty } \alpha \, d\alpha \int _{0}^{\infty } f(\beta )J_{\nu }(\alpha \beta )J_{\nu }(x\alpha )\beta \, d\beta .

Problem 5960

Prove that the integral equation

u(x) = \lambda \int _{0}^{\infty } J_{\nu }(xt)tu(t) \, dt, \quad \nu > -\frac{1}{2}

for $\lambda = \pm 1$ has the following solution:

u(x) = \frac{x^{\nu }}{(a^{2}+x^{2})^{\nu +\frac{1}{2}}} \pm \frac{\sqrt{\pi }x^{\nu -1}e^{-ax}}{2^{\nu }\Gamma \left(\nu +\frac{1}{2}\right)},

where $a$ denotes an arbitrary parameter, subject to the sole condition $\mathrm{Re}\left(a\right) > 0$ .