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In your answers below, for the variable $\lambda$ type the word lambda, for $\gamma$ type the word gamma; otherwise treat these as you would any other variable.

We will solve the heat equation $\displaystyle u_t = 4\, u_{xx}, \quad 0< x < 4, \quad t \ge 0$ with boundary/initial conditions: $\displaystyle \begin{aligned} u_x(0,t) &= 0,\\ u(4,t) &= 0, \end{aligned} \quad \mathrm{and} \quad u(x,0) = \begin{cases} 0, & 0 < x < 2 \\ 4, & 2 \le x < 4 \end{cases}$

This models temperature in a thin rod of length $L = 4$ with thermal diffusivity $\alpha = 4$ where one end is insulated and the other end has fixed temperature and the initial temperature distribution is $u(x,0)$ .
For extra practice we will solve this problem from scratch.

Separate Variables.

Separate variables.
Assume $u(x,t) = X(x)\,T(t)$ and split the PDE into two differential equations, one with $X$ and one with $T$ .
= = $-\lambda$
(Notation: Write X'' and T ' for derivatives. Place all constants in the differential equation with T).

DE for $X(x)$ : $= 0$
Boundary conditions for $X(x)$ :

(Enter boundary equations: e.g. " $X'(0)=10$ ")
DE for $T(t)$ : $= 0$

Find Eigenfunctions for $X(x)$.

The problem splits into cases based on the sign of $\lambda$ .
(Notation: For the cases below, use constants a and b)

Case 1: $\lambda = 0$
$X(x) =$
Plugging the boundary values into this formula gives
$0 = X'(0) =$
$0 = X(4) =$
So $X(x) =$
We can ignore this case.
Case 2: $\lambda = -\gamma^2 < 0$ (In your answers below use gamma instead of lambda)
$X(x) =$
Plugging the boundary values into this formula gives
$0 = X'(0) =$
$0 = X(4) =$
So $X(x) =$
We can ingore this case.
Case 3: $\lambda = \gamma^2 > 0$ (In your answers below use gamma instead of lambda)
$X(x) =$
Plugging the boundary values into this formula gives
$0 = X'(0) =$
$0 = X(4) =$

Which leads us to the eigenvalues $\lambda_n = \gamma_n^2$ where $\gamma_n =$
and eigenfunctions $X_n(x) =$
(Notation: Eigenfunctions should not include any constants a or b. Here it is convenient to let $n=0$ correspond to the smallest eigenvalue, $n=1$ to the second smallest, and so on.)

Solve for $T(t)$.

Plug the eigenvalues $\lambda_n = \gamma_n^2$ from Case 3 into the differential equation for $T(t)$ and solve:
$T_n(t) =$
(Notation: use c for the unknown constant.)

Combining all of the $X_n$ and $T_n$ we get that
$u(x,t)=\displaystyle\sum\limits_{n=0}^\infty \;C_n$
where $C_n$ are unknown constants.

Eigenfunction expansion.

We compute $C_n$ by plugging $t=0$ into the formula for $u(x,t)$ and setting equal to the initial heat distribution given in the problem.
$u(x,0) = \displaystyle \sum\limits_{n=0}^\infty C_n$ $= \begin{cases} 0, & 0 < x < 2 \\ 4, & 2 \le x < 4 \end{cases}$

So the $C_n$ are coefficients in the eigenfunction expansion of $u(x,0)$ and they are computed just like the standard Fourier coefficients.
$\displaystyle C_n=\frac{2}{4}\int_{2}^{4}$ $\hskip 3pt dx$

$\hskip 16pt =$

You can earn partial credit on this problem.