Toronto Math Forum

Taking partial fourier transform gives us:   
 $\hat{u}_{xx}-k^2\hat{u}=0$ giving us an ODE, which then can be solved: $\hat{u}=A(k)e^{-|k|x}+B(k)e^{|k|y}$
   
The second term is discarded since we want a finite solution, otherwise it would violate boundary condition 3.

So now we take $\hat{u}_{x} = -|k|A(k)e^{-|k|x} = 0 =  \hat{h}$

What is $\hat{h}$

We know that $h(y) = -g'(y)$ and $g(y) = \frac{2}{y^2 + 1}$ with fourier transform of $e^{-|k|}$. We then use a property of the fourier transform, where $\hat{g} =  ik\hat{f}$ implies = $g = f'(x)$
Thus:

$\hat{h} =  -ik\hat{g}$

Thus
$ -|k|A(k) = -ik e^{-|k|}$, rearranging gives us:

$A(k) = \frac{ike^{-|k|}}{|k|} $

And $u(x,k) = \frac{ike^{-|k|}}{|k|}e^{-|k|x}  $

We can put this in a fourier integral, and integrate to get u(x,y):

$u(x,y) = \int_{-\infty}^{\infty} \frac{ike^{-|k|}}{|k|}e^{-|k|x} e^{iky} dk$
Split into two integrals, to simplify the process:

$u(x,y) = \int_{0}^{\infty} i e^{-k(1+x - iy)} + \int_{-\infty}^{0} -ie^{k(1+x+iy)} = \frac{(i)((1+x - iy) - (1+x+iy))}{(1+x)^2 + y^2} = \frac{2y}{(1+x)^2 + y^2}$

We let $u = X(x)T(t)$
then plug in:

$T''X - X''T + 4XT = 0 $

The boundary conditions imply:

$X(0) =  X(\pi) = 0 $

and

$u(x,0) = f(x)$
$u_t (x,0) = x^2 - \pi x$

Using same strategy as we always do with a separation of variables, divide by u, and move the $-4$ over to the other side.


$\frac{T''}{T}  - \frac{X''}{X} = -4$

Since all three terms are constants, set $\frac{X''}{X} =  -\lambda$ and $\frac{T''}{T} = -\lambda - 4$

Solving these ODEs:

$X(x) =  Acos\sqrt{\lambda x} + B\sin \sqrt{\lambda x}$
and plugging in the conditions for X will imply that $A = 0$
$X(\pi) = B\sin \sqrt{\lambda \pi} = 0$

and in order to have an eigenfunction with a nontrivial solution, $\sqrt{\lambda} \pi = \pi n$ where n is an integer. Therefore eigenfunction of $n^2$.

Finally, for T: (I dropped a minus sign on the exam, see my comment at the bottom)

$T = C\sin\sqrt{\lambda + 4} t + D\cos \sqrt{\lambda + 4}t $ becomes

$T =  C\sin\sqrt{n^2 + 4} t + D\cos \sqrt{n^2 + 4}t$

Then our general solution ought to have the form

$u(x,t) =  \sum_{n=1}^{\infty}(C_n\sin\sqrt{n^2 + 4} t + D_n\cos \sqrt{n^2 +4}t)\sin(nx)$

Apply the BC of $u(x,0) = 0$

$0 =  \sum_{n=1}^{\infty}(C_n\sin\sqrt{n^2 + 4} (0) + D_n\cos \sqrt{n^2 + 4}(0)\sin(nx)$

Second term's coefficient has to be 0 satisfy this BC:

$u_t(x,t) = \sum_{n=1}^{\infty}(\sqrt{n^2+4}(-C_n\cos\sqrt{n^2 + 4}(t) )\sin(nx)$

Then for the last condition:

$x^2 - \pi x = \sum_{n=1}^{\infty}(\sqrt{n^2+4}(-C_n )\sin(nx)$

Now to find the coefficient:

$C_n = \frac{-2}{\pi \sqrt{n^2+4}}\int_{0}^{\pi}\sin(nx)(x^2 - \pi x) dx$

Split this into two integrals,

$ \frac{-2}{\pi \sqrt{n^2+4}}(\int_{0}^{\pi}x^2\sin(nx - \int_{0}^{\pi} \sin(nx)\pi x)$

Integrate by parts to get

$ = (\frac{-2}{\pi \sqrt{n^2+4}})(\frac{-\pi^2 \cos \pi}{n} - \frac{2}{n^3}(1-\cos(n\pi)) + \frac{\pi^2 \cos \pi n}{n} =   \frac{4}{(\pi \sqrt{n^2+4}) n^3}(1-\cos(n\pi))$

Final series:

$u = \sum_{n=1}^{\infty} \frac{8}{(\pi \sqrt{n^2+4}) n^3})\sin(nx)(\cos(\sqrt{n^2+4})t)$

for odd n.

EDIT: After careful revision, I realized I must have dropped a minus sign on the final. The correct solution should be involving (n^2+4) not (n^2-4), and the above steps have been corrected to reflect that:

@Jingxuan and @George, thank you for pointing out my mistake!

Given a Lagrangian of $$ L(q,\dot{q})  = \frac{m\dot{\vec{q}}^2}{2} + eA( \vec {q})\dot{\vec{q}} - V(q) $$ and a Hamiltonian of $$ H (q,p) =  \frac{m}{2} \sum_{i =1}^{n} (\frac{p_i - eA(q)_i)}{m})^2  + V(q) $$

To find $ \ddot{q}$:

Use the Euler Lagrange equations: $ \frac{\partial L}{\partial \vec{q}} -  \frac{d}{dt} \frac{\partial L}{\partial \dot{\vec{q}}} = 0 $, we get:

$m\dot{q_j} + e\frac{\partial A}{\partial q_j} - m\ddot{q_j} - e (\frac{\partial A_j}{\partial q_i} \dot{q_j} - e \frac{\partial A_j}{\partial t})  = 0$ then $$ m\ddot{q_j} = m\dot{q_j} + e(\frac{\partial A_j}{\partial q_j}  - \frac{\partial A_j}{\partial q_i} \dot{q_j} - \frac{\partial A_j}{\partial t} ) $$

$$  \ddot{q_j}  =  \frac{1}{m} (m\dot{q_j} + e(\frac{\partial A_j}{\partial q_j} - \frac{\partial A_j}{\partial q_i} \dot{q_j} - \frac{\partial A_j}{\partial t} )) $$

Alternatively, one can use Hamilton's equations of motion to find $\dot{q}, \ddot{q}$ ,

$$ \frac{\partial H}{\partial p}  = \dot{q}  =  - \frac{2p}{2m} + \frac{eA}{2m} $$, where taking the time derivative gives us:

$$ \ddot{q}  = - \frac{\dot{p}}{m} + \frac{e}{2m} \frac{\partial A }{\partial q} \dot{q} $$

We start by taking the $ u = X(x)Y(y) $, then plugging in gives us:

$$ X'' Y + Y'' X  = -\lambda XY $$

$$ X(0)Y(y) = X'(a)Y(y) = 0  \ \ and \ \ X(x)Y(0) = X(x)Y(b)' = 0 $$

Dividing both of these expressions by $XY$ gives us

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

$$ X(0) = X'(a) = 0  \ \ and \ \ Y(0) = Y(b)' = 0 $$

Now we know that both $\frac{Y''}{Y} $ and $\frac{X''}{X} $ are independent of each other, i.e. they should be equivalent to some constant. Introduce constants $\lambda_1, \lambda_2$ such that $ \lambda = \lambda_1 + \lambda_2$, thus  $\frac{Y''}{Y} = -\lambda_{1} $ and $\frac{X''}{X} = -\lambda_{2} $

Then, we can examine the different cases of $\lambda_{1,2}$.

i) If both $\lambda_{1} = 0 = \lambda_{2} $:

We get a simplified eigenvalue problem of:
 $$ X'' = 0 , Y'' = 0 $$

Meaning that:

$$ X = A_0 x + B_0 , Y = C_0y + D_0  $$

Running it through the boundary conditions, we can easily show that: $ B_0 = 0 , D_0 = 0 , A_0 = 0 , C_0 = 0 $
I.e. this leads to a trivial solution of the eigenvalue problem.

For $\lambda_{1}, \lambda_{2} >0 $

We will get eigenvalue problem of $$X'' + \lambda_2 X = 0 , Y'' + \lambda_1  Y = 0$$

This results in:

$$X(x) = Acos\sqrt{\lambda_2}x + Bsin\sqrt{\lambda_2}x$$
$$Y(y) = Ccos\sqrt{\lambda_1}y + Dsin\sqrt{\lambda_1}y$$

Apply the boundary conditions, and we get:
$ A = 0 , C= 0, 0 =  \sqrt{\lambda_2}Bcos\sqrt{\lambda_2}a, 0 = \sqrt{\lambda_1}Dcos\sqrt{\lambda_1}b$

We're in search of nontrivial solutions, which can be attained if $\sqrt{\lambda_{1,2}}b,a =  \frac{\pi(2n+1)}{2} $, thus we have eigenvalues and eigenfunctions of:

$$ \lambda_1 =  (\frac{\pi(2m+1)}{2b})^2 , Y_{m} = sin(\frac{\pi(2m+1)}{2b})y , \lambda_2 = (\frac{\pi(2n+1)}{2a})^2, X_{n} =  sin(\frac{\pi(2n+1)}{2a}x) $$

For the case of $\lambda_1 , \lambda_2 < 0 $ we solve the eigenvalue problem of:

$$X''  - \lambda_2 X = 0 , Y'' - \lambda_1 Y = 0 $$, which gives us, in turn:

$$X(x) =  Ae^{\sqrt{\lambda_2} x} + Be^{-\sqrt{\lambda_2} x} , Y(y) = Ce^{\sqrt{\lambda_1} y} + De^{-\sqrt{\lambda_1} y} $$

Apply the boundary conditions to get: $ A+ B = 0, C+D = 0$ , $ 0 =  \sqrt{\lambda_2} A (e^{\sqrt{\lambda_2}a} +e^{-\sqrt{\lambda_2}a}) = 2A\sqrt{\lambda_2}\cosh(\sqrt{\lambda_2}a) $

and $ 0 = \sqrt{\lambda_1}C(e^{\sqrt{\lambda_1}b} +e^{-\sqrt{\lambda_1}b})  = 2C\sqrt{\lambda_1}\cosh(\sqrt{\lambda_1}b) $

But since the $\cosh$ function never reaches 0, we can't have a nontrivial solution. Therefore there only exists a trivial solution in this case.

Note: updated solution to reflect feedback

Let $\a > 0 $ , find the fourier transforms of $$ (x^2 + a^2)^{-1} $$ and $$ x(x^2 + a^2)^{-1} $$. HINT: Consider the fourier transform of $e^{-\beta|k|} $.

Starting with the hint, we can take:

$ f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{ikx} e^{-\beta|k|} dk = \frac{1}{\sqrt{2\pi}}[ \int_{0}^{\infty} e^{-k(\beta - ix)} dk + \int_{-\infty}^{0} e^{k(\beta + ix)} dk] $

Integrating and evaluating at the bounds will leave us:

$ \frac{1}{\sqrt{2\pi}} [\frac{1}{\beta + ix} + \frac{1}{\beta - ix} ] = \frac{2\beta}{x^2 + \beta^2} \times \frac{1}{\sqrt{2\pi}} $

Then, for problem 1:

let $\beta = a $

Then using the property that $\hat{f} = F $ and $\hat{F} = f $, we can note:

the fourier transform of the above is merely $\frac{1}{x^2 + a^2} \ times \frac{2a}{\sqrt{2\pi}} $. The above property then implies that some multiple of the function $e^{-a|k|}$ is our desired function for the fourier transform, specifically:

$g(k) = \frac{\sqrt{2\pi}}{2a} e^{-a|k|}$ 's fourier transform, $\hat{g(x)} = \frac{\sqrt{2\pi}}{2a} \ times \frac{2\beta}{x^2 + a^2} \times \frac{1}{\sqrt{2\pi}}  = \frac{1}{x^2 + a^2}$ as desired.

For the second function's fourier transform: note that $g(x) = xf(x) \rightarrow \hat{g(k)} =  i\hat{f'(k)} $ is a property, and that the second function is x times the previous problem's function.

Thus, take the derivative of $\hat{f(k)} $ i.e. $\frac{d}{dk} ( e^{-a|k|}) =  \frac{-ak e^{-a|k|}}{|k|} $ So then:

the forward fourier transform is:

$\frac{-i\sqrt{\pi}k e^{-a|k|}}{\sqrt{2}|k|}$

a) We have that given $ u_t + xtu_x = xte^{-t^2}{2} $

We use: $\frac{dx}{tx} = \frac{dt}{1} = \frac{du}{xte^{\frac{-t^2}{2}}} $

Integrating gives us the relation:

$ c + \frac{t^2}{2}  = x $   WRONG. V.I.

This gives us the following characteristic curves:


b) Finding the general solution:

We plug in our value for x into:

$\frac{dt}{1} = \frac{du}{(\frac{t^3 e^{\frac{-t^2}{2}}+ Cte^{\frac{-t^2}{2}}} $|


Integration yields:

$$ u(x,t) = D +\frac{ Ce^{-t^2 / 2} - \frac{Ce^{-t^2 / 2} (t^2 + 1)}{3}}{4}$$

$$ D = \phi(C') = u - \frac{ Ce^{-t^2 / 2} - \frac{Ce^{-t^2 / 2} (t^2 + 1)}{3}}{4} $$

$$ u = \phi( x - \frac{t^2}{2}) + \frac{ ( x - \frac{t^2}{2})e^{-t^2 / 2} - \frac{( x - \frac{t^2}{2})e^{-t^2 / 2} (t^2 + 1)}{3}}{4} $$

Giving us a lovely general solution.


c) Solving (1) with the initial condition of $ u(x,0) = x $ :

$$ u(x,0) = \phi(x) + \frac{x}{3} - \frac{x}{12} = x$$

Therefore $\phi(x) = \frac{3/4} x $

Giving us a solution:

$$ u(x,t) = \frac{3( x - \frac{t^2}{2})}{4}+ \frac{ ( x - \frac{t^2}{2})e^{-t^2 / 2} - \frac{( x - \frac{t^2}{2})e^{-t^2 / 2} (t^2 + 1)}{3}}{4} $$

This solution can be split into two parts: for $y> 0 $ and $ y< 0 $, yielding

$$e^{-x}$$
$$e^{x} $$ respectively.

So u is split over those two regions. For the first one, we have

$$ u(x,t) = \frac{1}{\sqrt{4\pi t}} \int_0^{\infty} \exp(\frac{-(x-y)^2) - 4yt}{4t}) dy $$

This calls for completing the square, then integrating as usual:

$$ u(x,t) = \frac{1}{\sqrt{4\pi t}} \int_0^{\infty} \exp(\frac{-(x-y)^2) - 4yt}{4t}) $$

$$ u(x,t) =  \frac{1}{\sqrt{4\pi t}}  \int_0^{\infty} \exp(\frac{-(y-(x-2t)^2)}{4t}  exp(t-x) dy $$

$$ u(x,t) = \frac{exp(t-x)}{\sqrt{4\pi t}} \int_0^{\infty} \exp(\frac{-(y-(x-2t)^2)}{4t} $$

Now use the error function $ \erf(z)  = \frac{2}{\sqrt{\pi}} \int_{0}^{z} e^{-z^2} dz $ to answer:
let $ z = (y -(x-2t)) / \sqrt{4t} $ and $dz = \frac{dy}{\sqrt{4t}} $ giving us the integral:

$$ u(x,t) = \frac{\sqrt{4t} exp(t-x)}{\sqrt{4\pi t}}\int_{\frac{2t-x}{\sqrt{4t}}}^{\infty} e^{-z^2}dz $$

$$ u(x,t) = \frac{exp(t-x)}{2} erf(\infty) - \frac{\exp(t-x)}{2}\erf(\frac{2t-x}{\sqrt{4t}}) $$

Repeating these steps for the other region with $\phi(x) = e^x $ gives us:

$$ u(x,t) = -\frac{exp(t+x)}{2} \erf(-\infty) + \frac{\exp(t+x)}{2}erf(\frac{-2t-x}{\sqrt{4t}}) $$
\
Note that error function at $\pm \infty = \pm 1 $

Adding these together should give the solutions that Jingxuan posted.