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Exercises 7.5 Exercise
1.
Using the transformations \(p=x+vt\) and \(q=x-vt\text{,}\) solve the following,
\begin{equation*}
\frac{\partial^{2} u}{\partial t^{2}} = v^{2}\frac{\partial^{2} u}{\partial x^{2}};
\end{equation*}
\begin{equation*}
\left.\frac{\partial u}{\partial t}\right\vert_{t=0};\quad u(x,0) = f(x).
\end{equation*}
Answer . \(u(x,t) = \sum\limits_{n=1}^{\infty}b_{n}\cos \frac{n\pi vt}{l}\sin \frac{n\pi x}{l}\)
2. Find the solution of the equation of a vibrating string of length \(l\) satisfying the initial conditions \(y=f(x)\) and \(\frac{\partial y}{\partial t}=g(x)\) when \(t=0\text{.}\)
It is assumed that the equation of a vibrating string is
\begin{equation*}
\frac{\partial^{2} u}{\partial t^{2}} = v^{2}\frac{\partial^{2} u}{\partial x^{2}}.
\end{equation*}
Answer .
\begin{equation*}
y(x,t)=\sum\limits_{n=1}^{\infty}\left[b_{n}\cos \frac{n\pi vt}{l}+c_{n}\sin \frac{n\pi vt}{l}\right]\sin \frac{n\pi x}{l}.
\end{equation*}
\begin{equation*}
b_{n}=\frac{2}{l}\int\limits_{n=1}^{l}f(x)\sin \frac{n\pi x}{l}\,dx
\end{equation*}
and
\begin{equation*}
c_{n}=\frac{2}{n\pi v}\int\limits_{n=1}^{l}g(x)\sin \frac{n\pi x}{l}\,dx.
\end{equation*}
3.
A tightly stretched string of length \(l\) fastened at both ends, is disturbed from the position of equilibrium by imparting to each of its points an initial velocity of magnitude \(f(x)\text{.}\) Show that the solution of the problem is
\begin{equation*}
u(x,t) = \frac{2}{l}\sum\limits_{n=1}^{\infty}\left[\int\limits_{n=1}^{l}f(x)\sin \frac{n\pi x}{l}\,dx\right]\sin \frac{n\pi x}{l}\cos \frac{n\pi vt}{l}.
\end{equation*}
4.
Find the solution of the equation
\begin{equation*}
\frac{\partial^{2} y}{\partial x^{2}} = \frac{1}{v^{2}}\frac{\partial^{2} y}{\partial t^{2}}
\end{equation*}
subject to the boundary conditions \(y(0,t)=0=y(l,t); y(x,0)=\phi(x)\) and
\begin{equation*}
\left.\frac{\partial y}{\partial t}\right\vert_{t=0}=\psi(x).
\end{equation*}
Answer .
\begin{equation*}
y=\phi(x) \cos \frac{n\pi vt}{l}+\frac{l\psi(x)}{n\pi v}\frac{\sin \frac{n\pi vt}{l}}{\frac{n\pi x}{l}}.
\end{equation*}
5.
The vibration of an elastic string of length \(l\) are governed by the one - dimensional wave equation
\begin{equation*}
\frac{\partial^{2} u}{\partial t^{2}} = v^{2}\frac{\partial^{2} u}{\partial x^{2}}.
\end{equation*}
The string is fixed at the ends, \(u(0,t) =0=u(l,t)\) for all \(t\text{.}\) The initial deflection is
\begin{equation*}
u(x,0)=\begin{cases}
x, & \text{for} \quad 0\leq x\leq l/2\\
l-x, & \text{for} \quad l/2\leq x\leq l
\end{cases}
\end{equation*}
and the initial velocity is zero. Find the deflection of the string at any instant of time.
Answer .
\begin{equation*}
u=\sum\limits_{n=1}^{\infty}\left(\frac{4l}{n^{2}\pi^{2}}\sin \frac{n\pi}{2}\right)\sin \frac{n\pi x}{l}\cos \frac{n\pi vt}{l}
\end{equation*}
6.
Find the solution of
\begin{equation*}
\frac{\partial^{2}u}{\partial x^{2}}=h^{2}\frac{\partial u}{\partial t}
\end{equation*}
for which
\begin{equation*}
u(0,t) =0=u(\lambda,t), \quad u(x,0) =\sin\frac{\pi x}{\lambda}
\end{equation*}
by the method of variables separable.
Answer .
\begin{equation*}
u=\sin\frac{n\pi}{\lambda}e^{-\pi^{2}t/h^{2}\lambda^{2}}.
\end{equation*}
7.
Solve
\begin{equation*}
\frac{\partial^{2}u}{\partial x^{2}}=\frac{\partial u}{\partial t}
\end{equation*}
when \(u=0\) for \(t \to \infty\) and \(x=0\) or, \(\lambda \text{.}\)
Answer .
\begin{equation*}
u(x,t) = \sum\limits_{n=1}^{\infty}b_{n}\sin \frac{n\pi x}{\lambda}e^{-n^{2}\pi^{2}t/\lambda^{2}}
\end{equation*}
8.
Solve
\begin{equation*}
\frac{\partial u}{\partial t}= h^{2} \frac{\partial^{2}u}{\partial x^{2}},
0 \lt x \lt \pi, t \lt 0.
\end{equation*}
under the boundary conditions
\begin{equation*}
\left.\frac{\partial u}{\partial x}\right\vert_{(0,t)}=0=\left.\frac{\partial u}{\partial x}\right\vert_{(\pi,t)},
\end{equation*}
and
\begin{equation*}
\left.\frac{\partial u}{\partial x}\right\vert^{(x,0)}=\sin x.
\end{equation*}
Answer .
\begin{equation*}
u(x,t) = \frac{2}{\pi}-\frac{4}{\pi}\sum\limits_{m=1}^{\infty}(4m^{2}-1)^{-1}e^{-4m^{2}h^{2}t}\cos 2mx.
\end{equation*}
9.
Find the temperature \(u(x,t)\) in a bar of length \(\lambda\) perfactly insulated and whose ends are kept at temp. zero while the initial temp. is given by
\begin{equation*}
f(x) = \begin{cases}
x, & 0 \lt x \lt \lambda/2\\
\lambda-x, & \lambda/2 \lt x \lt \lambda
\end{cases}
\end{equation*}
Answer .
\begin{equation*}
u(x,t) = \frac{4\lambda}{\pi^{2}}\left[\frac{1}{1^{1}\sin\frac{\pi x}{\lambda}}e^{-\pi^{2}h^{2}t/\lambda^{2}}+\frac{1}{3^{1}\sin\frac{3\pi x}{\lambda}}e^{-3^{2}\pi^{2}h^{2}t/\lambda^{2}}+\cdots \right]
\end{equation*}
10.
The heat flow in a bar of length 10 cm of homogeneous material is governed by the partial differential equation
\begin{equation*}
\frac{\partial u}{\partial t}= h^{2} \frac{\partial^{2}u}{\partial x^{2}}.
\end{equation*}
The ends of the bar are kept at temperature \(0^{o} C\text{,}\) and the initial temperature is \(f(x) =x(10-x)\text{.}\) Find the temperature in the bar at any instant of time.
Answer .
\begin{equation*}
u=\frac{800}{\pi^{2}}\sum\limits_{m=1}^{\infty}\frac{e^{-n^{2}\pi^{2}h^{2}t/100}}{n^{2}}\sin\frac{n\pi x}{10}
\end{equation*}
where \(n\) is odd.
11.
Heat flows in a semi - infinite rectangular plate, the end \(x=0\) is kept at temp. \(T^{o} C\) and the edges \(y=0\) and \(y=a\) at temp. zero, then show that the temp. at any point \((x,y) \) is given by
\begin{equation*}
u(x,y) = \frac{4T}{\pi}\sum\limits_{r=1}^{\infty}\frac{1}{2^{r+1}} \sin\frac{(2r+1)\pi y}{a}e^{-\frac{(2r+1)\pi x}{a}}
\end{equation*}
12.
A square plate has its faces and its edges \(x=0\) and \(x=\pi\) for \((0 \lt y \lt \pi)\) is insulated. Its edges \(y=0\) and \(y=\pi\) are kept at temperatures zero and \(f(x)\) respectively. Show that the steady temperature is given by
\begin{equation*}
u(x,y) = \frac{1}{2\pi}a_{o}y+\sum\limits_{n=1}^{\infty}a_{n}\frac{\sinh ny}{\sinh n\pi} \cos nx
\end{equation*}
where
\begin{equation*}
a_{n}=\frac{2}{\pi}\int\limits_{0}^{\pi}f(x) \sin\cos nx\,dx, \quad n=0,1,2,3,\cdots.
\end{equation*}
13.
Solve
\begin{equation*}
\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}} = 0
\end{equation*}
for \(0 \lt x \lt \pi, \quad 0 \lt y \lt \pi\) under the boundary conditions \(u(x,0)=x^{2}, u(x,\pi)=0,\)
\begin{equation*}
u_{x}(0,y)=\frac{\partial}{\partial x}u_{x}(0,y)=0=u_{x}(\pi,y)
\end{equation*}
Answer .
\begin{equation*}
u(x,y)=\frac{\pi}{3}(\pi-y)+4\sum\limits_{n=1}^{\infty}(-1)^{n}\frac{\sinh(\pi-y)\cos nx}{n^{2}\sinh n\pi}
\end{equation*}
14.
The boundary of a semi - circular plate of radius \(a\) is kept at temperature
\begin{equation*}
u(a,\theta) = k\theta(\pi-\theta),
\end{equation*}
while the diameter of the plate is maintained at \(0^{o} C\text{.}\) Find the steady state temperature distribution \(u(r,\theta) \) of the plate assuming the lateral surfaces of the plate to be insulated.
Answer .
\begin{equation*}
u(r,\theta)= 8\sum\limits_{n=1}^{\infty}(\frac{r}{a})^{2n-1}\frac{\sin(2n-1)\theta}{(2n-1)^{3}\sinh n\pi}
\end{equation*}
