Toronto Math Forum

By separation of variables, we get $u(x,y) = X(x) Y(y)$. Plugging this in $ u_{xx} +u_{yy}=0$ we get
 $$\frac{X_{xx}}{X} +\frac{Y_{yy}}{Y}=0  \implies \\ \begin{align}&\frac{X_{xx}}{X}=-\lambda,\label{5-4}\\ &\frac{Y_{yy}}{Y}=\lambda \label{5-5} \end{align}$$ where $\lambda \in R$.
From the boundary conditions given, we must have
$$\begin{align}&X(0) = 0\\
&X(\pi) = 0\end{align}$$ in order to have a nontrivial solution for u(x,y).
Equations (2)-(4) are the associated eigenvalue problem.


We know $$X(x) = A \sin(\sqrt\lambda x)+B\cos(\sqrt\lambda x)$$ but we rule out the cosine term because of the first boundary condition on x ( equation 3). Equation 4 implies $$X(\pi) = A\sin(\sqrt\lambda \pi)= 0 \implies \sqrt\lambda \pi = n \pi, n \in N \implies \lambda = n^2 \implies X(x) = A sin(nx)$$ where $\sin(nx)$ are the X eigenfunctions and $-n^2$ is the eigenvalue.

From equation (2), we get $$Y(y) = C e^{\sqrt\lambda y} +De^{-\sqrt\lambda y} $$ but we rule out $C e^{\sqrt\lambda y}$ because this reaches infinity as y approaches infinity. We are left with $$Y(y)=De^{-n y}$$ where $e^{-ny}$ are the y eigenfunctions and $n^2$ is the eigenvalue.

So $$u(x,y) = X(x) Y(y) = \sum_{n = 1}^{\infty} C_n sin(nx) e^{-n y}$$

The boundary condition $$u_y(x,0)=\cos(x) \implies cos(x) =\sum_{n = 1}^{\infty} -n C_n sin(nx) $$ This is a sine fourier series with L = $\pi$, we call $A_n = -n C_n$, and solve for $A_n$:

$$A_n=\frac{2}{\pi} \int_0^{\pi} \cos(x) \sin(nx) dx=\frac{2}{2 \pi} \int_0^{\pi} \Bigl(\sin(n-1)x + \sin(n+1)x \Bigr)\;dx \tag{$\checkmark$}\\ \\ = \frac{-1}{\pi} \left.\frac{\cos(n-1)x}{n-1} \right|_{0}^{\pi} + \left.\frac{\cos(n+1)x)}{n+1} \right|_{0}^{\pi} \implies$$


$A_n = 0$           n is odd
$A_n = \frac{4n}{\pi (n^2-1)}$    n is even

So
$C_n = 0$                  n is odd
$C_n=-\frac{4}{\pi(n^2-1)}$        n is even

So $$ u(x,y) = \sum_{n = 1, \;n \;even}^{\infty} -\frac{4}{\pi (n^2-1)}  \sin(nx) e^{-n y} \;for \;y>0, 0<x<\pi$$ which can be written as
$$u(x,y) = \sum_{m = 1}^{\infty} -\frac{4}{\pi \Bigl((2m)^2-1 \Bigr)}  \sin(2mx) e^{-2m y}, \;\; for \;y>0, 0<x<\pi$$


Edit: my final answer has remained the same, except it's written in a different form (I'm not sure what error I've made after Prof. Ivrii's checkmark, perhaps someone can point it out). I've also formatted the latex a bit differently

Problem 4 of the 2012 final exam: http://forum.math.toronto.edu/index.php?topic=177.0
It asks to prove that the total energy (kinetic + potential) is constant with time. I get up to $$\frac{d}{dt} k(t) + p(t) = u_x(\infty) u_t(\infty) - u_x(-\infty) u_t(-\infty) $$ How do I prove $$u_x(\infty) u_t(\infty) - u_x(-\infty) u_t(-\infty) = 0 $$ using the boundary conditions? right now I know $u_t =0$ only when t = 0 (and x is large)

The website says HA 1-10 is a good preparation. Does that mean the final exam doesn't cover chapter 10? (i.e  variational methods)

Does 3c want us to go through steps a and b in spherical coordinates, or just write u as a function of r, theta, and phi?

link: http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter8/S8.P.html     (3a)

Since P(x,y,z) is a polynomial of degree 0, it is a constant. So $U=x^2+y^2+z^2-c_0 (x^2+y^2+z^2)$, but you can't write this as a sum of homogenous harmonic polynomials since there is a term c_0 remaining?

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter4/S4.5.P.html

For problem 6f, I get $b_n=0$, which I know is wrong.
I have $$b_n=\frac{2}{\pi} \int_0 ^{\pi} \sin((m-1/2)x) \sin((n+1/2)x) dx \\
=1/pi \int_0 ^{\pi} \cos((m-n-1)x) -\cos((m+n)x) dx =0$$
Separately for $n \neq m-1$ and $n=m-1$. Why am I getting b_n=0?

Problem 1c asks to investigate how many negative eigenvalues there are:
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter4/S4.2.P.html#mjx-eqn-a

I understand that we have the hyperbola $$\alpha + \beta+ \alpha \beta l=0$$ which divides the $(\alpha,\beta)$ plane into three zones, as he problem states. But how does that actually help us find the number of negative eigenvalues?