Question

3. Using separation of variables to solve the heat equation, u- kuxx on the interval 0 < x< 1 with boundary conditions ux(0, t) = 0 and ux(1, t) yields the general solution, 1, 0<x < 1/2 0, 1/2 x<1 Determine the coefficients An (n = 0, 1, 2, . . .) when u(x,0) = f(x) =

Answer 1

Lets us write in variable separable form the solution as

a(x,t) = g(x)h(t)

Substituting back into the equation

0 u

g(x)}'(t) = kg" (x)h(t)

where

dg dt

The boundary conditions thus become

g (0)0

g(1) 0

Similarly the initial conditions are

a(x, 0) = g(x)h(0) = f(z)

Looking at the modified heat equation &

rearranging we have,

g"(x) h'(t) g(r) = kh(1)

Since the LHS is dependent upon x, only & RHS dependent on t only that means both must be equalling a constant say $-\lambda^2$

The minus sign is chosen so as to get a periodic solution satisfying periodic neumann conditions at all times.

Hence we have two sets of ordinary differential equations, viz.

$g''(x) + \lambda^2g(x) =0$

$h'(t)+ k\lambda^2h(t)=0$

The general solution for the first one is

$g(x) = c_1\cos{\lambda x} +c_2\sin{\lambda x}$

where $c_1$ & $c_2$ are constants satisfying boundary conditions. Now,

$g'(x) = -\lambda\left(c_1 \sin{\lambda x} + c_2 \cos{\lambda x}\right )$

$g'(0) = 0 \implies c_2 = 0$

$g'(1)=0 \implies \sin{\lambda } = 0 \implies \lambda = n\pi \quad n=0,\pm 1,\pm 2,\pm 3, \ldots$

Since for each value of n we can find a different $\lambda_n$ with $g_n(x)$ & $h_n(t)$

The $h_n(t)$ satisfies the same equation as h(x)

$h_n'(t)+ k\lambda_n^2h_n(t)=0$

the solution is obtained by multiplying $e^{k\lambda^2t}$ on both sides of equation as in

$e^{k\lambda_n^2t}h_n'(t)+ e^{k\lambda_n^2t}k\lambda_n^2h_n(t)=0 \implies \frac{d\left(e^{k\lambda_n^2t}h_n \right )}{dt} = 0$

Thus one can write

$h_n = C_ne^{-k\lambda_n^2t} \quad n=0, \pm 1, \pm 2, \pm 3, \ldots$

with $C_n$ as an arbitrary constant to be determined.

Hence the complete solution is of the form

$u(x,t) = \sum_{n=-\infty}^{\infty}g_n(x)h_n(t) = \sum_{n=-\infty}^{\infty}B_n \cos{\left(n\pi x \right )}e^{-k\lambda_n^2t}$, wherein

$B_n$ absorbs $c_1C_n$

For the above can be further simplified using the property that cosine is even function with respect to the sign of n

Thus we have

$u(x,t) = \sum_{n=0}^{\infty}A_n \cos{\left(n\pi x \right )}e^{-k\lambda_n^2t}$

where

$A_i = B_i + B_{-i} \quad i> 0$

Since the above is a fourier series we can find $A_n$ as

$A_n = 2\int_{0}^{1}f(x)\cos{\left(n\pi x \right )}dx$

Now given that

$f(x) = 1 \quad \forall x \in \left[0\ \frac{1}{2}\right]$

$f(x) =0 \quad \forall x \in \left[ \frac{1}{2}\ 1\right]$

$\implies A_n = 2\int_{0}^{1/2}\cos{\left(n\pi x \right )}dx$

For

$n=0 \implies A_0 = 2\int_{0}^{1/2}dx = 1$

For $n>0 \implies A_n = 2\int_{0}^{1/2}\cos{n\pi x}dx = 2\frac{\sin{n\pi x}}{n \pi}\big|_{x=0}^{x=1/2} = \frac{2}{n \pi}\sin{\frac{n\pi}{2}}$

Moreover, $\lambda_0 = 0$

Hence final solution

is of form

u(z, t) = 1 +