Question

Find the temperature in a bar of length l with both ends insulated and with initial temperature in the rod being sin πx/L.

Answer 1

Answer on Question 66341 - 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)

We look for a specific type of solution; namely, a product of a function of $t$ only and a function of $x$ only:

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

We substitute this function $v$ 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 this equality depends only upon $t$. The right-hand side is independent of $t$. It follows that

$\frac{X^{\prime\prime}(x)}{X(x)}=-\lambda,$

where $\lambda$ is a constant. Then

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

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

$T^{\prime}(t)+\lambda a^{2}T(t)=0,\qquad t\in(0,+\infty)$ (4)

and the boundary value problem

$X^{\prime\prime}(x)+\lambda X(x)=0,\quad x\in(0,L),$ (5)

$X^{\prime}(0)=0,\quad X^{\prime}(L)=0.$ (6)

respectively. Since we wish to have $u_{x}=0$ for $x=0$ and $x=L$, we only consider those solutions of the last equation which also satisfy conditions (6).

This homogeneous problem always has the trivial solution $X=0$, but this is of no use to us. We are interested in cases where this is not the only solution. It is possible for non-negative $\lambda$ only.

If $\lambda=0$, then there exists a nonzero solution $X_{0}=1$ of (5), (6). For $\lambda>0$ we have $X(x)=C_{1}\cos\sqrt{\lambda}x+C_{2}\sin\sqrt{\lambda}x$. Substituting $X$ into (6) 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 (4) 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,

$A_{0}=\frac{1}{L}\int_{0}^{L}\sin\frac{\pi x}{L}\,dx,\qquad A_{n}=\frac{2}{L}\int_{0}^{L}\sin\frac{\pi x}{L}\cos\frac{\pi nx}{L}\,dx.$

$A_{0}=\int_{0}^{L}\sin\frac{\pi x}{L}dx=\frac{L}{\pi}\int_{0}^{L}\sin\frac{\pi x}{L}d\left(\frac{\pi x}{L}\right)=-\frac{L}{\pi}\left.\cos\frac{\pi x}{L}\right|_{0}^{L}=\frac{2L}{\pi}.$

3

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

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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