Math Problems Vault

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.

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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}.

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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.

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Remark.

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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.

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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.

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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).

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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).

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

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Remark.

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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).

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Remark.

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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.

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Remark.

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

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Remark.

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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}

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Remark.

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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 $x = 0$ .

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Remark.

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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).

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Remark.

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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.

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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 $t > 0$ .

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Remark.

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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

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