For the general solution of the differential equation in the following cases use a and b for your constants and list the functions in alphabetical order, for example $y = a\cos (x) + b \sin(x)$. For the variable $\lambda$ type the word lambda, for $\alpha$ type alpha, otherwise treat these as you would any other variable.


Solve the heat equation $$\displaystyle k\frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}, 0< x < L, t > 0$$ $$\displaystyle\frac{\partial u}{\partial x}(0,t) = 0, \frac{\partial u}{\partial x}(L,t) = 0, t > 0$$ $$\displaystyle u(x,0) = \begin{cases} 1, &0 < x < \frac{L}{2}\\ 0, &\frac{L}{2}<x<L\\ \end{cases}$$

This models the heat equation in a long thin rod with insulated ends. Since we have not solved this boundary value problem, we have to start from scratch
We look for solutions of the form $u(x,t) = X(x)T(t)$.
The PDE can be rewritten using this solution as (placing constants in the DE for T) into
= = $-\lambda$

Note: Use the prime notation for derivatives, so the derivative of $X$ is written as $X^\prime$. Do NOT use $X^\prime(x)$
The boundary conditions in the PDE translate into boundary conditions for the differential equation in X:
Since these differential equations are independent of each other, they can be separated
DE in X: $= 0$
BCs in X: Note: You have to input all four answers to this question or WeBWorK will mark it wrong.

DE in T: $= 0$

This leads us to a Sturm-Louiville problem. In each case the general solution in X is written with constants a and b and the general solution in Y is written with constants c and d.
Case 1: $\lambda = 0$
$X(x) =$
Using this formula

Making X a constant function. We get the constant solution another way, so we will discard this case.

Case 2: $\lambda = -\alpha^2$
$X(x) =$
$X^\prime(0) =$ $= 0$
That simplifies X down to . Using the simplified X:
$X^\prime(L) =$ $= 0$
Therefore $a =$ and

$X =$

Case 3: $\lambda = \alpha^2$
$X(x) =$
$X^\prime(0) =$ $= 0$
So X simplifies to . Using the simplified X:
$X^\prime(L) =$ $= 0$
Which leads us to the eigenvalues $\alpha =$
Note n = 0 is the constant from above, including it would have been redundant.
Since this case has nontrivial solutions we now solve the differential equation in T using c as the constant:
$T =$
So for each n the product $u = XT =$ (without constants)is a solution of the PDE with boundary conditions:
$\displaystyle k\frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}, 0< x < L, t > 0$ $\displaystyle\frac{\partial u}{\partial x}(0,t) = 0, \frac{\partial u}{\partial x}(L,t) = 0, t > 0$

To get the initial conditions we need a series solution of the form
$u(x,t)=\displaystyle\sum\limits_{n=0}^\infty A_n$
$u(x,0)=\displaystyle\sum\limits_{n=0}^\infty A_n$$= \begin{cases} 1, &0 < x < \frac{L}{2}\\ 0, &\frac{L}{2}<x<L\\ \end{cases}$
which we recognize as a Fourier series:
$\displaystyle A_n=\frac{2}{L}\int_0^{\frac{L}{2}}$ $\hskip 3pt dx$

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