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:

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- Jun 19th 2007, 01:33 PM #1degomanGuest

- Jun 19th 2007, 02:15 PM #2

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

By definition,

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

By definition,

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

- Jun 19th 2007, 02:23 PM #3

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

Then,

$\displaystyle u_t = ae^{at}\sin bx$

$\displaystyle u_x = be^{at}\cos bx$

$\displaystyle u_{xx} = -b^2 e^{at}\sin bx$

Thus,

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

Make each factor equal to zero:

$\displaystyle 1) \ \ \ a+b^2 = 0 \implies a=-b^2$

$\displaystyle 2) \ \ \ e^{at} = 0 \mbox{ cannot be since }e^{at}>0$

$\displaystyle 3) \ \ \ \sin bx = 0 \implies b = 2\pi n \mbox{ for }n\in \mathbb{Z}$

- Jun 19th 2007, 02:33 PM #4

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4)I think you made a mistake. That is not the solution to:

$\displaystyle \nabla^2 u = \frac{1}{k^2}u_t$

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 $\displaystyle \ln z = \ln |z| + i\arg(z) \mbox{ with }-\pi<\arg z \leq \pi$ is holomorphic on $\displaystyle \mathbb{C} - (-\infty,0]$. That means $\displaystyle \mbox{Re} \ln z \mbox{ and }\mbox{Im} \ln z$ are harmonic functions.

We can write,

$\displaystyle \ln |z| + i \arg(z) = \ln \sqrt{x^2+y^2} + i\left( \tan^{-1} \frac{y}{x} +k \right)$

That means the real part is $\displaystyle \ln \sqrt{x^2+y^2}$ solves Laplace's equation.

- Jun 19th 2007, 03:00 PM #5

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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.

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

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

Then by substituting this into the PDE we have:

$\displaystyle X''Y + XY'' = \lambda XY$

Divide through by $\displaystyle XY$ to get:

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

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

$\displaystyle 1) \ \ \ \frac{X''}{X} = - k \implies X'' + kX = 0$

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

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

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