1. ## help PDE's!

I'm not so sure how to type my problems out legibly on here so here's a document with the problems i need help with:

2. 1) $\mbox{div}(\mbox{grad} u) = \nabla^2 u$ where $u=u(x,y,z)$

By definition,
$\mbox{grad} u = \frac{\partial u}{\partial x}\bold{i} + \frac{\partial u}{\partial y}\bold{j} + \frac{\partial u}{\partial z}\bold{k}$

By definition,
$\mbox{div} \left( \frac{\partial u}{\partial x}\bold{i} + \frac{\partial u}{\partial y}\bold{j} + \frac{\partial u}{\partial z}\bold{k} \right) = \frac{\partial^2 u}{\partial x^2}+\frac{\partial u^2}{\partial y^2}+\frac{\partial u^2}{\partial z^2}$

3. $u(x,t) = e^{at}\sin bx$

Then,
$u_t = ae^{at}\sin bx$

$u_x = be^{at}\cos bx$
$u_{xx} = -b^2 e^{at}\sin bx$

Thus,
$u_t - u_xx = ae^{at}\sin bx +b^2 e^{at}\sin bx = (a+b^2)e^{at}\sin bx = 0$

Make each factor equal to zero:
$1) \ \ \ a+b^2 = 0 \implies a=-b^2$
$2) \ \ \ e^{at} = 0 \mbox{ cannot be since }e^{at}>0$
$3) \ \ \ \sin bx = 0 \implies b = 2\pi n \mbox{ for }n\in \mathbb{Z}$

4. 4)I think you made a mistake. That is not the solution to:
$\nabla^2 u = \frac{1}{k^2}u_t$

5)The boring what is to substitute that into the PDE and verify that it solves it, i.e. adds up to zero. A more exciting way is to use the Cauchy-Riemann equations. The prinipal logarithm $\ln z = \ln |z| + i\arg(z) \mbox{ with }-\pi<\arg z \leq \pi$ is holomorphic on $\mathbb{C} - (-\infty,0]$. That means $\mbox{Re} \ln z \mbox{ and }\mbox{Im} \ln z$ are harmonic functions.
We can write,
$\ln |z| + i \arg(z) = \ln \sqrt{x^2+y^2} + i\left( \tan^{-1} \frac{y}{x} +k \right)$
That means the real part is $\ln \sqrt{x^2+y^2}$ solves Laplace's equation.

5. 3)I am really not sure about this one, but I think I know what I am doing. First, in your problem you have the Dirichlet problem in 3 variables, is that a mistake? I will do it in two variables.

$\frac{\partial ^2 u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2} = \lambda u$

Seperate variables and write $u(x,y) = X(x)Y(y)$.

Then by substituting this into the PDE we have:
$X''Y + XY'' = \lambda XY$

Divide through by $XY$ to get:

$\frac{X''}{X} + \frac{Y''}{Y} = \lambda \implies \frac{X''}{X} = \lambda - \frac{Y''}{Y}$

And hence by the seperation of variables techinque we have a system of ODE:

$1) \ \ \ \frac{X''}{X} = - k \implies X'' + kX = 0$
$2) \ \ \ \lambda - \frac{Y''}{Y} = - k \implies Y'' -(k+\lambda) Y = 0$

Look at the first equation together with the boundary value problem $u(x,y) =0 \mbox{ on }\partial \Omega$ implies that $X(0) = 0 \mbox{ and }X(L)=0$ for $L>0$ means non-trivial solutions can only exist iff $k>0$, that is the solutions are sines and cosines.

Look at the second equation by similar reasoning $-(k+\lambda) > 0 \implies k + \lambda < 0$. So if $k>0$ (by the first equation) it must mean that $\lambda < 0$ for otherwise $k+\lambda >0$ and that will lead to trivial solutions to the two point boundary value problem.