Question

Find the temperature in a bar of length l with both ends insulated and with initial
temperature in the rod being sin ( π x / l ) .
.

Answer 1

Answer on Question 68975 - Math - Differential Equations

Find the temperature in a bar of length $L$ with both ends insulated and with initial temperature in the rod being $\sin\frac{\pi x}{L}$.

Solution: We consider the initial boundary value problem for the heat equation

$u_{t}=a^{2}u_{xx},\qquad x\in(0,L),\quad t>0,$ (1)

$u(0,x)=\sin\frac{\pi x}{L},$ (2)

$u_{x}(t,0)=u_{x}(t,L)=0.$ (3)

Let us find a special type of non-trivial solutions

$v(t,x)=T(t)X(x)$ (4)

of (1) that satisfies conditions (3). We will show that there exists a countable set of such solutions

$v_{n}(t,x)=T_{n}(t)X_{n}(x),\quad n=0,1,2,\ldots$

Then we will look for a solution $u$ of (1)–(3) in the form

$u(t,x)=\sum_{n=0}^{\infty}v_{n}(t,x).$

We substitute (4) into the differential equation (1), and divide by $a^{2}v$. This gives

$\frac{T^{\prime}(t)}{a^{2}T(t)}=\frac{X^{\prime\prime}(x)}{X(x)}.$

The left-hand side of the equality depends only upon $t$ and the right-hand one is independent of $t$. It follows that

$\frac{T^{\prime}(t)}{a^{2}T(t)}=\frac{X^{\prime\prime}(x)}{X(x)}=-\lambda,$

where $\lambda$ is a constant. Thus $v(t,x)=T(t)X(x)$ is a solution of the heat equation that satisfies boundary conditions (3) if and only if $T$ and $X$ satisfy the ordinary differential equation

$T^{\prime}(t)+\lambda a^{2}T(t)=0,\qquad t\in(0,+\infty)$ (5)

and the boundary value problem

$X^{\prime\prime}(x)+\lambda X(x)=0,\quad x\in(0,L),$ (6)

$X^{\prime}(0)=0,\quad X^{\prime}(L)=0.$ (7)

respectively.

The Sturm-Liouville problem (6)–(7) always admits the trivial solution $X=0$, but this is of no use to us. We must find all values of $\lambda$

for which there exist non-trivial solutions of (6)–(7). It is possible for non-negative $\lambda$ only. In fact, if $\lambda<0$, then equation (6) has the general solution

$X(x)=C_{1}\sinh(\sqrt{-\lambda}x)+C_{2}\cosh(\sqrt{-\lambda}x).$

From boundary conditions (7) we have

$C_{1}=0,\quad C_{2}\sinh(\sqrt{-\lambda}\,L)=0.$

Therefore $\sinh(\sqrt{-\lambda}\,L)=0$, which is impossible for the positive number $\sqrt{-\lambda}\,L$.

If $\lambda=0$, then there exists a nonzero solution $X_{0}=1$ of (6), (7). For $\lambda>0$ we have $X(x)=C_{1}\cos\sqrt{\lambda}x+C_{2}\sin\sqrt{\lambda}x$. Substituting $X$ into (7) yields

$X^{\prime}(x)=-C_{1}\sqrt{\lambda}\sin\sqrt{\lambda}x+C_{2}\sqrt{\lambda}\cos\sqrt{\lambda}x;$

$X^{\prime}(0)=0\quad\Rightarrow\quad C_{2}=0;$

$X^{\prime}(L)=0\quad\Rightarrow\quad C_{1}\sqrt{\lambda}\sin\sqrt{\lambda}L=0\quad\Rightarrow\quad\sin\sqrt{\lambda}L=0,$

since both the constants $C_{1}$ and $C_{2}$ cannot be zero simultaneously. Then $X$ need not be identically zero if and only if $\sin\sqrt{\lambda}L=0$, that is, if

$\lambda_{n}=\frac{\pi^{2}n^{2}}{L^{2}},\qquad n=1,2,\ldots.$

These values are called the eigenvalues of the problem. The corresponding solutions are

$X_{n}(x)=\cos\frac{\pi nx}{L},\qquad n=1,2,\ldots.$

Next, we can solve equation (5) for all eigenvalues:

$T_{0}^{\prime}=0\quad\Rightarrow\quad T_{0}(t)=A_{0};$

$T_{n}^{\prime}+\frac{a^{2}\pi^{2}n^{2}}{L^{2}}T_{n}=0\quad\Rightarrow\quad T_{n}(t)=A_{n}e^{-\frac{a^{2}\pi^{2}n^{2}}{L^{2}}t}.$

We attempt to represent the solution $u$ of (1)–(3) as an infinite series

$u(t,x)=A_{0}+\sum_{n=1}^{\infty}A_{n}e^{-\frac{a^{2}\pi^{2}n^{2}}{L^{2}}t}\cos\frac{\pi nx}{L}.$

We need to determine the coefficients $A_{n}$ in such a way that initial condition (2) holds. We have

$u(0,x)=A_{0}+\sum_{n=1}^{\infty}A_{n}\cos\frac{\pi nx}{L}=\sin\frac{\pi x}{L}.$

The last series is called a Fourier series of function $\sin \frac{\pi x}{L}$. Moreover,

$$A _ {0} = \frac {1}{L} \int_ {0} ^ {L} \sin \frac {\pi x}{L} d x, \qquad A _ {n} = \frac {2}{L} \int_ {0} ^ {L} \sin \frac {\pi x}{L} \cos \frac {\pi n x}{L} d x.$$ $$A _ {0} = \int_ {0} ^ {L} \sin \frac {\pi x}{L} d x = \frac {L}{\pi} \int_ {0} ^ {L} \sin \frac {\pi x}{L} d \left(\frac {\pi x}{L}\right) = - \frac {L}{\pi} \cos \frac {\pi x}{L} \Big | _ {0} ^ {L} = \frac {2 L}{\pi}.$$ $$\begin{array}{l} A _ {n} = \int_ {0} ^ {L} \sin \frac {\pi x}{L} \cos \frac {\pi n x}{L} d x = \frac {1}{2} \int_ {0} ^ {L} \left(\sin \frac {\pi (n + 1) x}{L} + \sin \frac {\pi (n - 1) x}{L}\right) d x \\ = \frac {1}{2} \left(\frac {L}{\pi (n + 1)} \int_ {0} ^ {L} \sin \frac {\pi (n + 1) x}{L} d \left(\frac {\pi (n + 1) x}{L}\right) \right. \\ + \frac {L}{\pi (n - 1)} \int_ {0} ^ {L} \sin \frac {\pi (n - 1) x}{L} d \left(\frac {\pi (n - 1) x}{L}\right) \\ = \frac {1}{2} \left(- \frac {L}{\pi (n + 1)} \cos \frac {\pi (n + 1) x}{L} \Bigg | _ {0} ^ {L} - \frac {L}{\pi (n - 1)} \cos \frac {\pi (n - 1) x}{L} \Bigg | _ {0} ^ {L}\right) \\ = (1 - (- 1) ^ {n + 1}) \left(\frac {L}{\pi (n + 1)} + \frac {L}{\pi (n - 1)}\right) = \frac {2 (1 - (- 1) ^ {n + 1}) n L}{\pi (n ^ {2} - 1)} \\ = \left\{ \begin{array}{l l} \frac {8 k L}{\pi (4 k ^ {2} - 1)} & \text {if} \quad n = 2 k \\ 0 & \text {if} \quad n = 2 k - 1 \end{array} \right.. \\ \end{array}$$

Answer:

$$u (t, x) = \frac {2 L}{\pi} + \sum_ {k = 1} ^ {\infty} \frac {8 k L}{\pi (4 k ^ {2} - 1)} e ^ {- \frac {4 a ^ {2} \pi^ {2} k ^ {2}}{L ^ {2}} t} \cos \frac {2 \pi k x}{L}.$$

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