Past Exams

Consider the PDE

$\displaystyle u(x,y)\frac{\partial^2}{\partial x\partial y} u(x,y) = \frac{\partial}{\partial x} u(x,y) \frac{\partial}{\partial y} u(x,y).$

Show that the solution of this PDE is

$\displaystyle u(x,y)= f(x)g(y)$

for any differentiable functions $f(x)$

and

Consider the Laplacian equation

$\displaystyle \frac{\partial^2}{\partial x^2} u(x,y) + \frac{\partial^2}{\partial y^2} u(x,y) =0.$

Verify by direct explicit calculation that

$\displaystyle u_n(x,y) = \sin( n x) \sinh( n y)$

is a solution for this equation. What are the conditions on constant $n$

should you impose?

Solve the PDE

$\displaystyle 2\frac{\partial}{\partial x} u(x,t) + 3\frac{\partial}{\partial t} u(x,t) =0.$

with the initial condition

$\displaystyle u(x,t=0)=\sin(x).$

Find the general solution of the PDE

$\displaystyle 4\frac{\partial^2}{\partial x \partial t } u(x,t) + 5 \frac{\partial}{\partial x} u(x,t) =0.$

Hint There is a substitution which simplifies this problem considerably.

Find the general solution of the PDE

$\displaystyle x \frac{\partial}{\partial x } u(x,y) + y \frac{\partial}{\partial y} u(x,y) =0.$

Solve the Partial Differential Equation

$\displaystyle \frac{\partial^2}{\partial x \partial t } u(x,t) - 3 \frac{\parti... ...}{\partial x \partial t } u(x,t) - 4 \frac{\partial^2}{\partial x^2} u(x,t) =0.$

Consider the solution $u(x,t)$

of the diffusion equation

$\displaystyle \frac{\partial}{\partial t} u(x,t) = \frac{\partial^2}{\partial x^2} u(x,y), u(0,t)=u(L,t)=0.$

The

is between $0$

and

$\displaystyle 0\le x\le L,$

and

is between $0$

and $\infty$ :

$\displaystyle 0\le t\le \infty.$

Let
$\displaystyle M(T) = \max\limits_{0\le x \le L, 0\le t \le T} u(x,t).$
Does increases or decreases as increases? Why?
Let
$\displaystyle m(T) = \min\limits_{0\le x \le L, 0\le t \le T} u(x,t).$
Does increases or decreases as increases? Why?

Solve the initial value problem for the diffusion equation:

$\displaystyle \frac{\partial}{\partial t} u(x,t) = \frac{\partial^2}{\partial x^2} u(x,y), \ \ u(x,t=0)=\phi(x), \ \ -\infty <x<\infty,$

(1)

where the $\phi(x) = 1$ for $x>0$

and $\phi(x) = 3$ for $x<0$

. Write your solution in terms of the Error function

$\displaystyle {\rm {Erf}}(x) =\frac{2}{\sqrt{\pi}}\int\limits_0^x e^{-s^2} d s.$

Let

satisfy the diffusion equation

$\displaystyle \frac{\partial}{\partial t} u(x,t) = \frac{1}{2}\frac{\partial^2}{\partial x^2} u(x,y),$

Define the function $v(x,t)$

$\displaystyle v(x,t) = \frac{1}{\sqrt{t}} e^{\frac{x^2}{2 t}} u(\frac{x}{t},\frac{1}{t}).$

Show that

satisfies the “backward” diffusion equation

$\displaystyle \frac{\partial}{\partial t} u(x,t) =- \frac{1}{2}\frac{\partial^2}{\partial x^2} u(x,y),$

Consider waves in the resistant medium described by the wave-like PDE

$\displaystyle \frac{\partial^2}{\partial t^2} u(x,t) = \frac{\partial^2}{\partial x^2} u(x,y) - r \frac{\partial}{\partial t} u(x,y),$
$\displaystyle u(x=0,t) = u(x=L,t) = 0, u(x,t=0)=\phi(x), \frac{\partial}{\partial x} u(x,t=0) = \psi(x). \nonumber$

Find the general solution in form of a series for this problem if

$\displaystyle \frac{2 \pi}{L}<r<\frac{4\pi}{L}.$

Consider the diffusion where the material evaporates. The equation is given by

$\displaystyle \frac{\partial}{\partial t} u(x,t) = \frac{\partial^2}{\partial x^2} u(x,y) - - u(x,y),$
$\displaystyle u(x=0,t) = u(x=L,t) = 0, u(x,t=0)=\phi(x). \nonumber$

Find the general solution in form of a series for this problem.

Solve the Schroedinger equation

$\displaystyle \frac{\partial}{\partial t} u(x,t) = i \frac{\partial^2}{\partial x^2} u(x,y) -$
$\displaystyle u(x=0,t) = u(x=L,t) = 0, u(x,t=0)=\phi(x). \nonumber$