Section 15.5

The Fourier Method

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Problem 5846

For a homogeneous string of length ll, fastened at the ends x=0x = 0 and x=lx = 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: ∂2u∂x2=1a2∂2u∂t2\frac{\partial^{2} u}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial^{2} u}{\partial t^{2}}, a=T0/ρa = \sqrt{T_{0}/\rho }).

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Remark.
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Hint. The problem reduces to integrating the equation

∂2u∂x2=1a2∂2u∂t2,a=T0ρ \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∣x=0=0u\big|_{x=0} = 0, u∣x=l=0u\big|_{x=l} = 0 for every tt; u∣t=0=f(x)u\big|_{t=0} = f(x), ∂u∂t∣t=0=F(x)\dfrac {\partial u}{\partial t}\Big|_{t=0} = F(x) for 0<x<l0 < x < l.

Problem 5847

Study by means of the Fourier method the longitudinal vibrations of a cylindrical rod of length ll, one end of which is fixed while the other is free (the equation of small longitudinal vibrations of the rod --- problem 5798: ∂2u∂x2=1a2∂2u∂t2\frac{\partial^{2} u}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial^{2} u}{\partial t^{2}}, a=E/ρa = \sqrt{E/\rho }).

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Remark.
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Hint. The problem reduces to integrating the equation

∂2u∂x2=1a2∂2u∂t2,a=Eρ \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∣x=0=0u\big|_{x=0} = 0, ∂u∂x∣x=l=0\dfrac {\partial u}{\partial x}\Big|_{x=l} = 0 for every tt; u∣t=0=f(x)u\big|_{t=0} = f(x), ∂u∂t∣t=0=F(x)\dfrac {\partial u}{\partial t}\Big|_{t=0} = F(x) for 0<x<l0 < x < l.

Problem 5848

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

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Problem 5849

Integrate by the Fourier method the equation of torsional vibrations of a shaft ∂2θ∂x2=1a2∂2θ∂t2\dfrac {\partial^{2} \theta }{\partial x^{2}} = \dfrac {1}{a^{2}}\dfrac {\partial^{2} \theta }{\partial t^{2}}, a=GI/ka = \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).

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Remark.
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Hint. The integral of the vibration equation satisfies the conditions

θ∣x=0=0,∂2θ∂t2∣x=l=−c2∂θ∂x∣x=l \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 tt, and θ∣t=0=f(x)\theta \big|_{t=0} = f(x), ∂θ∂t∣t=0=F(x)\dfrac {\partial \theta }{\partial t}\Big|_{t=0} = F(x) for 0<x<l0 < x < l, where c=GI/k1c = \sqrt{GI/k_{1}} and k1k_{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.

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Remark.
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Hint. The boundary conditions have the form

∂2θ∂t2∣x=0=c2∂θ∂x∣x=0;∂2θ∂t2∣x=l=−c2∂θ∂x∣x=l,wherec=GIk1. \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=lx = l of a homogeneous rod, fastened at the point x=0x = 0, a weight of PP kg is suspended. Study the forced longitudinal vibrations of the rod, assuming that it is acted on by an external continuously distributed force ρY(x,t)\rho Y(x,t).

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Remark.
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Hint. The problem reduces to integrating the equation

∂2u∂t2=a2∂2u∂x2+Y(x,t),a=Eρ \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∣x=0=0;∂2u∂t2∣x=l=−c2∂u∂x∣x=l,c=gEP; 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∣t=0=0;∂u∂t∣t=0=0; u\big|_{t=0} = 0; \quad \frac{\partial u}{\partial t}\Big|_{t=0} = 0;

gg --- the acceleration of free fall.

Problem 5852

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

Fig. 15.15Fig. 15.15

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Problem 5853

A string of length ll is fastened at the ends x=0x = 0 and x=lx = l and at the initial moment of time, at the point x=cx = 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)u(x,t) of the string (see problem 5846).

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

Problem 5854

A perfectly insulated electric wire of length ll, at a certain moment of time, taken as the initial one, is charged to a certain constant potential p0p_{0}. Determine the magnitude of the voltage and the current ii after a time tt 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: ∂p∂x+L∂i∂t+Ri=0\frac{\partial p}{\partial x}+L\frac{\partial i}{\partial t}+Ri = 0, ∂i∂x+C∂p∂t+Gp=0\frac{\partial i}{\partial x}+C\frac{\partial p}{\partial t}+Gp = 0).

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Remark.
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Hint. The problem reduces to integrating the system of equations

∂p∂x+L∂i∂t+Ri=0,∂i∂x+C∂p∂t=0 \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∣x=0=0p\big|_{x=0} = 0, i∣x=l=0i\big|_{x=l} = 0 for every tt, p∣t=0=p0p\big|_{t=0} = p_{0}, i∣t=0=0i\big|_{t=0} = 0 for 0<x<l0 < x < l.

Problem 5855

A homogeneous string of length ll, under the influence of a strong tension T0T_{0}, is in a rectilinear position of equilibrium and is fastened at the ends x=0x = 0 and x=lx = l. At the point x=cx = c a concentrated harmonic force ρP(t)=ρAsin⁡ωt\rho P(t) = \rho A\sin \omega t acts on the string.

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

Fig. 15.16Fig. 15.16

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Remark.
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Hint. The problem can be reduced to integrating the equations

∂2u1∂x2=1a2∂2u1∂t2,∂2u2∂x2=1a2∂2u2∂t2,a=T0ρ \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:

u1∣x=0=0,u1∣x=c=u2∣x=c,u2∣x=l=0. u_{1}\big|_{x=0} = 0, \quad u_{1}\big|_{x=c} = u_{2}\big|_{x=c}, \quad u_{2}\big|_{x=l} = 0. T0∂u2∂x∣x=c−T0∂u1∂x∣x=c+Pρ=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 ll, fastened at the ends x=0x = 0 and x=lx = l, vibrates under the action of an external disturbing force ρY(x,t)\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.

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Remark.
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Hint. The equation of forced vibrations has the form

∂2u∂t2=a2∂2u∂x2+Y(x,t);a=T0ρ; \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=u1+u2u = u_{1}+u_{2}, where u1u_{1} is the integral of the homogeneous equation ∂2u∂t2=a2∂2u∂x2\dfrac {\partial^{2} u}{\partial t^{2}} = a^{2}\dfrac {\partial^{2} u}{\partial x^{2}}, satisfying the conditions:

u1∣x=0=0,u1∣x=l=0,u1∣t=0=f(x),∂u1∂t∣t=0=F(x), 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 u2u_{2} is the integral of the inhomogeneous equation, subject to the conditions:

u2∣x=0=0,u2∣x=l=0,u2∣t=0=0,∂u2∂t∣t=0=0. 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=0x = 0 and subject at the end x=lx = l to the action of a disturbing harmonic force, causing a displacement equal to Asin⁡ωtA\sin \omega t.

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Problem 5858

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

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

Problem 5859

Two dissimilar strings of length l1l_{1} and l2l_{2} are joined together at the point x=l1x = l_{1} and fastened at the ends x=0x = 0 and x=l1+l2x = l_{1}+l_{2}. Determine the period of transverse vibrations.

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Remark.
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Hint. The problem reduces to integrating the equations

∂2u1∂t2=a12∂2u1∂x2;∂2u2∂t2=a22∂2u2∂x2;a1=T0ρ1;a2=T0ρ2 \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

u1∣x=0=0,u2∣x=l1+l2=0,u1∣x=l1=u2∣x=l1; 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}}; ∂u1∂x∣x=l1=∂u2∂x∣x=l1. \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=l1AB = l_{1} and BC=l2BC = l_{2}, with a ball of mass MM placed at the point BB. Prove that the period TT of transverse vibrations is determined from the equation

1a1ctg⁡2πl1Ta1+1a2ctg⁡2πl2Ta2=M2πTT0, \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 T0T_{0} --- the magnitude of the tension of the string.

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Remark.
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Hint. One should take into account the condition

M∂2u1∂t2∣x=l1=M∂2u2∂t2∣x=l1=−T0∂u1∂x∣x=l1+T0∂u2∂x∣x=l1. 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 ACAC, fastened at the beginning AA and free at the end CC, consists of two inhomogeneous parts AB=l1AB = l_{1} and BC=l2BC = l_{2}. Prove that the period TT of longitudinal vibrations of the rod is determined by the equation

tg⁡2πl1a1T⋅tg⁡2πl2a2T=E1a2E2a1, \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 a1=E1ρ1a_{1} = \sqrt{\dfrac {E_{1}}{\rho_{1}}}, a2=E2ρ2a_{2} = \sqrt{\dfrac {E_{2}}{\rho_{2}}}.

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

∂2u1∂t2=a12∂2u1∂x2;∂2u2∂t2=a22∂2u2∂x2 \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

u1∣x=l1=u2∣x=l1;E1∂u1∂x∣x=l1=E2∂u2∂x∣x=l1. 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 ll, is suspended by the end x=0x = 0, while the end x=lx = l is left free. Study the forced vibrations of the rod.

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Remark.
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Hint. The problem reduces to integrating the equation

∂2u∂t2=a2∂2u∂x2+g \frac{\partial ^{2} u}{\partial t^{2}} = a^{2}\frac{\partial ^{2} u}{\partial x^{2}}+g

under the conditions: u∣t=0=0u\big|_{t=0} = 0; ∂u∂t∣t=0=0\dfrac {\partial u}{\partial t}\Big|_{t=0} = 0, u∣x=0=0u\big|_{x=0} = 0; ∂u∂x∣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 R1R_{1} and R2R_{2}, R1<R2R_{1} < R_{2}, can occur harmonically, with the period of the tone expressed by the formula

T1=2π(R2−R1)μρ0γp0, T_{1} = \frac{2\pi (R_{2}-R_{1})}{\mu }\sqrt{\frac{\rho _{0}}{\gamma p_{0}}},

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

tg⁡μ=μpμ2+1;p=R1R2(R2−R1)2; \tg \mu = \frac{\mu }{p\mu ^{2}+1}; \quad p = \frac{R_{1}R_{2}}{(R_{2}-R_{1})^{2}};

here γ\gamma --- Poisson's constant, p0p_{0} and ρ0\rho_{0} --- the initial pressure and density.

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Problem 5864

Prove that the equation of radial vibrations of a gas ∂2Φ∂r2+2r∂Φ∂r=1a2∂2Φ∂t2\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 Φ=Arcos⁡ωa(at−r)\Phi = \dfrac {A}{r}\cos \dfrac {\omega }{a}(at-r), which vanishes as r→∞r\to \infty.

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Problem 5865

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

∂2Φ∂r2+2r∂Φ∂r+1r2sin⁡θ∂∂θ(sin⁡θ∂Φ∂q)=1a2∂2Φ∂t2, \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

Φ=Acos⁡θr(cos⁡ωa(at−r)+aωrsin⁡ωa(at−r)), \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→∞r\to \infty.

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Problem 5866

Prove that the solution of problem 5802 (problem 5802: a perfectly flexible string of length ll is suspended by the end x=lx = l, and to the other end x=0x = 0 a weight of PP kg is attached; the density of the string ρ=A/l1+x\rho = A/\sqrt{l_{1}+x}, where M=2Al1M = 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=∑k=1∞(akcos⁡aλkt+bksin⁡aλkt)J0(μkxl), 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

ak=1lJ12(μk)∫0lf(x)J0(μkxl) dx;f(x)=u∣t=0, 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}, bk=1aμklJ12(μk)∫01F(x)J0(μkxl) dx,F(x)=∂u∂t∣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}, λk=μk24l−(ωa)2, \lambda _{k} = \sqrt{\frac{\mu _{k}^{2}}{4l}-\left(\frac{\omega }{a}\right)^{2}},

and μ1,μ2,…\mu_{1}, \mu_{2}, \ldots denote the positive roots of the equation J0(x)=0J_{0}(x) = 0.

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Problem 5867

The linear density of a suspended string varies according to the law ρ=axm\rho = ax^{m}, where m>−1m > -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=∑k=1∞NkJm(μkxl)xm2sin⁡(12gl(m−1)μkt+φk), 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 μ1,μ2,…\mu_{1}, \mu_{2}, \ldots --- the positive roots of the equation Jm(x)=0J_{m}(x) = 0, and NkN_{k} and φk\varphi_{k} --- constants depending on the initial conditions.

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Problem 5868

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

ρY=ρAsin⁡ωt, \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=Aω2(J0(ωar)J0(ωaR)−1)sin⁡ωt, 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 RR --- the radius of the membrane.

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Problem 5869

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

Determine its temperature at the moment of time tt (the heat conduction equation of the rod --- problem 5812: ∂2u∂x2=1a2∂u∂t\frac{\partial^{2} u}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial u}{\partial t}, a=k/(cρ)a = \sqrt{k/(c\rho )}).

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Remark.
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Hint. The problem reduces to integrating the heat conduction equation ∂u∂t=a2∂2u∂x2\dfrac {\partial u}{\partial t} = a^{2}\dfrac {\partial^{2} u}{\partial x^{2}}, a=k/(cρ)a = \sqrt{k/(c\rho )} under the conditions: u∣x=0=0u\big|_{x=0} = 0, u∣x=l=0u\big|_{x=l} = 0, for every tt; u∣t=0=f(x)u\big|_{t=0} = f(x) for 0<x<l0 < x < l.

Problem 5870

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

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Remark.
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Hint. The second of the boundary conditions of the preceding problem should be replaced by the condition: k∂u∂x∣x=l=−hu∣x=lk\dfrac {\partial u}{\partial x}\Big|_{x=l} = -hu\big|_{x=l}, where kk and hh 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.

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Remark.
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Hint. The boundary conditions have the form

∂u∂x∣x=0=hku∣x=0,∂u∂x∣x=l=−hku∣x=l. \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: 1R2∂2u∂θ2−bu=1a2∂u∂t\frac{1}{R^{2}}\frac{\partial^{2} u}{\partial \theta^{2}}-bu = \frac{1}{a^{2}}\frac{\partial u}{\partial t}, a=k/(cρ)a = \sqrt{k/(c\rho )}, b=hl/(σk)b = hl/(\sigma k)).

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Remark.
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Hint. The problem reduces to integrating the equation

∂v∂t=a2∂2v∂θ2,u=ve−bt \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∣t=0=f(θ) for 0<θ<2π. u\big|_{t=0} = f(\theta ) \ \text{for} \ 0 < \theta < 2\pi .
Problem 5873

Given a homogeneous sphere of radius RR, 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 rr 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°0°. Determine the temperature of any point inside the sphere at the moment of time tt (the equation of radial heat propagation in the sphere --- problem 5816: ∂2v∂r2=1a2∂v∂t\frac{\partial^{2} v}{\partial r^{2}} = \frac{1}{a^{2}}\frac{\partial v}{\partial t}, v(r,t)=ru(r,t)v(r,t) = ru(r,t), a=k/(cρ)a = \sqrt{k/(c\rho )}).

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Remark.
?

Hint. The problem reduces to integrating the equation ∂v∂t=a2∂2v∂r2\dfrac {\partial v}{\partial t} = a^{2}\dfrac {\partial^{2} v}{\partial r^{2}}, u=vru = \dfrac {v}{r}, a=k/(cρ)a = \sqrt{k/(c\rho )} and to satisfying the conditions: v∣r=0=0v\big|_{r=0} = 0, v∣r=R=0v\big|_{r=R} = 0 for every tt; v∣t=0=rf(r)v\big|_{t=0} = rf(r) for 0<r<R0 < 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°0°.

?
Remark.
?

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

∂v∂r∣r=R+(H−1R)v∣r=R=0,H=hk. \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 RR. 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

∂2u∂x2+∂2u∂y2+∂2u∂z2=1a2∂2u∂t2, \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=γp0/ρ0a = \sqrt{\gamma p_{0}/\rho_{0}}, where p0p_{0} and ρ0\rho_{0} --- the initial pressure and density, and γ\gamma --- Poisson's constant.

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

∂u∂r∣r=R=0 for every t,u∣t=0=f(r),∂u∂t∣t=0=F(r) \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<R0 < r < R.

Problem 5876

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

Fig. 15.17Fig. 15.17

?
Remark.
?

Hint. The problem reduces to integrating the equation x∂2u∂x2+∂u∂x=1a2∂2u∂t2x\dfrac {\partial^{2} u}{\partial x^{2}}+\dfrac {\partial u}{\partial x} = \dfrac {1}{a^{2}}\dfrac {\partial^{2} u}{\partial t^{2}}, a=ga = \sqrt{g} under the conditions: u∣x=0=u\big|_{x=0} = a finite value, u∣x=l=0u\big|_{x=l} = 0 for every tt, u∣t=0=f(x)u\big|_{t=0} = f(x), ∂u∂t∣t=0=0\dfrac {\partial u}{\partial t}\Big|_{t=0} = 0 for 0<x<l0 < x < l.

Problem 5877

Given a cylindrical tube of radius RR, 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 rr, φ\varphi, zz, directing the axis zz along the axis of the tube, and then one should seek the integral of the equation

∂2u∂r2+1r∂u∂r=1a2∂2u∂t2, \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∣r=0=u\big|_{r=0} = a finite value, ∂u∂r∣r=R=0\dfrac {\partial u}{\partial r}\Big|_{r=R} = 0 for every tt; u∣t=0=f(r)u\big|_{t=0} = f(r), ∂u∂t∣t=0=F(r)\dfrac {\partial u}{\partial t}\Big|_{t=0} = F(r) for 0<r<R0 < 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 RR and fastened at the rim (the equation of small vibrations of the membrane --- problem 5811: ∂2u∂x2+∂2u∂y2=1a2∂2u∂t2\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=T0/σa = \sqrt{T_{0}/\sigma }).

?
Remark.
?

Hint. The problem reduces to integrating the equation

∂2u∂r2+1r∂u∂r=1a2∂2u∂t2, \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=T0/σa = \sqrt{T_{0}/\sigma } under the conditions u∣r=0=u\big|_{r=0} = a finite value, u∣r=R=0u\big|_{r=R} = 0 for every tt and u∣t=0=f(r)u\big|_{t=0} = f(r), ∂u∂t∣t=0=F(r)\dfrac {\partial u}{\partial t}\Big|_{t=0} = F(r) for 0<r<R0 < r < R.

Problem 5879

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

?
Remark.
?

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

∂u∂t=a2(∂2u∂r2+1r∂u∂r),a=kcρ. \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∣r=0=u\big|_{r=0} = a finite value, u∣r=R=0u\big|_{r=R} = 0 for every tt and u∣t=0=f(r)u\big|_{t=0} = f(r) for 0<r<R0 < r < R.

Problem 5880

A cylinder with base radius RR and height hh has, throughout the entire experiment, the temperature of its lower base and lateral surface equal to 0°0°, while the temperature of the upper base is a certain function of rr. 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 ∂2u∂r2+1r∂u∂r+∂2u∂z2=0\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∣r=0=u\big|_{r=0} = a finite value, u∣z=0=0u\big|_{z=0} = 0, u∣r=R=0u\big|_{r=R} = 0, u∣z=h=f(r)u\big|_{z=h} = f(r) for 0<r<R0 < 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: ∂u∂r∣r=R=0\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°0°.

?
Remark.
?

Hint. The boundary condition on the lateral surface has the form: ∂u∂r∣r=R+h1u∣r=R=0\dfrac {\partial u}{\partial r}\Big|_{r=R}+h_{1}u\big|_{r=R} = 0, where h1=hkh_{1} = \dfrac {h}{k}; for the meanings of the letters hh and kk see problem 5870 (problem 5870: kk and hh --- coefficients of internal and external thermal conductivity of the material of the rod).

Problem 5883

A cylinder with base radius RR and height hh has the temperature of both bases equal to 0°0°, while the temperature of the lateral surface represents a given function of zz: u=f(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 ll and mm (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°0°. It is required to determine the temperature of each point of the plate after a time tt from the initial moment (the heat conduction equation of the plate --- problem 5813: ∂2u∂x2+∂2u∂y2=1a2∂u∂t\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}} = \frac{1}{a^{2}}\frac{\partial u}{\partial t}, a=k/(cρ)a = \sqrt{k/(c\rho )}).

?
Remark.
?

Hint. The problem reduces to integrating the equation

∂u∂t=a2(∂2u∂x2+∂2u∂y2),a=kcρ \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 at x=0,u=0 at x=lfor 0<y<m and for every t, 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 at y=0,u=0 at y=mfor 0<x<l 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∣t=0=f(x,y)for 0<x<l and 0<y<m. 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 ll, mm, nn (three of its edges coinciding with the coordinate axes). Throughout the entire time of observation the lateral faces are held at temperature 0°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 ∂2u∂x2+∂2u∂y2+∂2u∂z2=0\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=0u = 0 at x=0x = 0, u=0u = 0 at x=lx = l, u=0u = 0 at y=0y = 0, u=0u = 0 at y=my = m, u=f(x,y)u = f(x,y) at z=0z = 0, u=F(x,y)u = F(x,y) at z=nz = n.

Problem 5886

Investigate the small vibrations of a membrane, fastened at the edges and having the shape of a rectangle, whose sides ll and mm coincide with the coordinate axes xx and yy (the equation of small vibrations of the membrane --- problem 5811: ∂2u∂x2+∂2u∂y2=1a2∂2u∂t2\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=T0/σa = \sqrt{T_{0}/\sigma }).

?
Remark.
?

Hint. The problem reduces to integrating the equation

∂2u∂x2+∂2u∂y2=1a2∂2u∂t2,a=T0σ \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 at x=0,u=0 at x=l, u = 0 \ \text{at} \ x = 0, \qquad u = 0 \ \text{at} \ x = l, u=0 at y=0,u=0 at y=m. u = 0 \ \text{at} \ y = 0, \qquad u = 0 \ \text{at} \ y = m. u∣t=0=f(x,y),∂u∂t∣t=0=F(x,y)for 0<x<l, 0<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: ∂2u∂x2=1a2∂u∂t\frac{\partial^{2} u}{\partial x^{2}} = \frac{1}{a^{2}}\frac{\partial u}{\partial t}, a=k/(cρ)a = \sqrt{k/(c\rho )}).

?
Remark.
?

Hint. The desired temperature satisfies the equation

∂u∂t=a2∂2u∂x2,a=kcρ \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∣t=0=f(x)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 tt from the initial moment.

?
Remark.
?

Hint. One should integrate the equation of heat propagation in the rod under the conditions: u∣x=0=φ(t)u\big|_{x=0} = \varphi (t), u∣t=0=f(x)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: ∂2u∂x2+∂2u∂y2=1a2∂u∂t\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}} = \frac{1}{a^{2}}\frac{\partial u}{\partial t}, a=k/(cρ)a = \sqrt{k/(c\rho )}).

?
Remark.
?

Hint. The desired temperature satisfies the equation

∂2u∂x2+∂2u∂y2=1a2∂u∂t,a=kcρ \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∣t=0=f(x,y)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∣x=0=φ(y,t);u∣t=0=f(x,y). 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)M(x,y,z) of unbounded space, assuming that the initial distribution of temperature is known.

?
Remark.
?

Hint. The desired temperature satisfies the equation

∂u∂t=a2(∂2u∂x2+∂2u∂y2+∂2u∂z2),a=kcρ \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∣t=0=f(x,y,z)u\big|_{t=0} = f(x,y,z).

Problem 5892

A flexible homogeneous string rotates with constant angular velocity ω\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)u(x,t) of the string after a time tt from the initial moment (the equation of vibrations of the string --- problem 5804: ∂∂x((l2−x2)∂u∂x)=1a2∂2u∂t2\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=ω/2a = \omega /\sqrt{2}).

?
Remark.
?

Hint. The problem reduces to integrating the equation

(l2−x2)∂2u∂x2−2x∂u∂x=1a2∂2u∂t2,a=ω2 (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∣x=0=0u\big|_{x=0} = 0, u∣x=l=u\big|_{x=l} = a finite value for every tt and u∣t=0=f(x)u\big|_{t=0} = f(x), dudt∣t=0=F(x)\dfrac {du}{dt}\Big|_{t=0} = F(x) for 0<x<l0 < x < l.

Problem 5893

Prove that Laplace's equation

∂2u∂r2+2r∂u∂r+1r2∂2u∂θ2+ctg⁡θr2∂u∂θ=0 \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=(Arm+Brm+1)Pm(cos⁡θ), u = \left(Ar^{m}+\frac{B}{r^{m+1}}\right)P_{m}(\cos \theta ),

where AA and BB denote arbitrary constants, mm is a nonnegative integer, and Pm(cos⁡θ)P_{m}(\cos \theta ) is the Legendre polynomial of index mm.

?
Problem 5894

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

f(r)=A0+A1r+A2r2+…+Anrn+…for r<R, f(r) = A_{0}+A_{1}r+A_{2}r^{2}+\ldots +A_{n}r^{n}+\ldots \quad \text{for } r < R, f(r)=B1r+B2r2+B3r3+…+Bn+1rn+1+…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: ∂2u∂r2+2r∂u∂r+1r2∂2u∂θ2+ctg⁡θr2∂u∂θ=0\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=(Arm+Brm+1)Pm(cos⁡θ)u = \left(Ar^{m}+\frac{B}{r^{m+1}}\right)P_{m}(\cos \theta )) as would satisfy the condition

u∣θ=0=f(r). u\big|_{\theta =0} = f(r).
?
Problem 5895

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

Fig. 15.18Fig. 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: ∂2u∂r2+2r∂u∂r+1r2∂2u∂θ2+ctg⁡θr2∂u∂θ=0\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=(Arm+Brm+1)Pm(cos⁡θ)u = \left(Ar^{m}+\frac{B}{r^{m+1}}\right)P_{m}(\cos \theta )) that, for r=Rr = R (on the surface of a sphere of radius RR centered at the origin), it turns into a given function of the angle θ\theta, i.e. u∣r=R=f(θ)u\big|_{r=R} = f(\theta ) for 0⩽θ⩽π0 \leqslant \theta \leqslant \pi.

?
Remark.
?

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

f(θ)=∑n=0∞AnPn(cos⁡θ), f(\theta ) = \sum _{n=0}^{\infty } A_{n}P_{n}(\cos \theta ), An=2n+12∫0πf(θ)Pn(cos⁡θ)sin⁡θ dθ. 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 RR is divided into two hemispheres, charged to potentials: v1v_{1} (upper) and v2v_{2} (lower). Find the potential v(r,θ)v(r,\theta ) of such a conductor at any point of the field (Fig. 15.19).

Fig. 15.19Fig. 15.19

?
Remark.
?

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

v∣r=R=f(θ)={v1for 0⩽θ⩽π2,v2for π2⩽θ⩽π. 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 RR with a soot-blackened surface is subject to the action of sunlight in air whose temperature equals 0°0°. Determine the steady-state temperature of the interior points of the ball (Fig. 15.20).

Fig. 15.20Fig. 15.20

?
Remark.
?

Hint. The problem reduces to integrating Laplace's equation (see problem 5893: ∂2u∂r2+2r∂u∂r+1r2∂2u∂θ2+ctg⁡θr2∂u∂θ=0\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:

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

where HH is constant, and

f(θ)={Mcos⁡θ,for 0⩽θ⩽π2(M=const);0,for π2⩽θ⩽π. 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 ∂4u∂x4+1a2∂2u∂t2=0\dfrac {\partial^{4} u}{\partial x^{4}}+\dfrac {1}{a^{2}}\dfrac {\partial^{2} u}{\partial t^{2}} = 0 under the conditions u=0u = 0, ∂u∂x=0\dfrac {\partial u}{\partial x} = 0 at x=0x = 0, x=lx = l, expressing that both ends of the beam are rigidly fastened (Fig. 15.21).

Fig. 15.21Fig. 15.21

?
Remark.
?

Hint. The characteristic numbers λ1,λ2,λ3,…\lambda_{1}, \lambda_{2}, \lambda_{3}, \ldots of the problem are determined from the equation

ch⁡λl⋅cos⁡λl=1, \ch \lambda l \cdot \cos \lambda l = 1,

and the corresponding fundamental functions

Xn(x)=(sh⁡λnl−sin⁡λnl)(ch⁡λnx−cos⁡λnx)−(ch⁡λnl−cos⁡λnl)(sh⁡λnx−sin⁡λnx) 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)(0,l).