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### Messages - Tristan Fraser

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1
##### Final Exam / Re: FE-P1
« on: April 11, 2018, 08:13:38 PM »
@Jingxuan: Thank you!

$(e^{-x} \psi(x)) = 4e^{x} + C$

and then:

$\psi(x) = 4e^{2x} + Ce^{x}$

Now, this gives us:

$u(x,t) = 4e^{2(x-4t)} + Ce^{x-4t} + e^{-2(x+4t)}$

Then checking continuity gives us (x=4t):

$u(x,t) = 4+ C +e^{-2(8t)} = 3+ e^{-16t}$ implying that $C= -1$.

$$u(x,t) = 4e^{2(x-4t)} - e^{x-4t} + e^{-2(x+4t)}\qquad 0<x<4t$$

2
##### Final Exam / Re: FE-P7
« on: April 11, 2018, 06:34:09 PM »
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:

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

3
##### Final Exam / Re: FE-P1
« on: April 11, 2018, 05:55:16 PM »
General solution of

$u = \phi (x+4t) + \psi (x-4t)$
$u_x = \phi'(x+4t) + \psi'(x-4t)$
$u_t = 4(\phi' (x+4t) - \psi' (x-4t)$
For $x>4t$:

$4e^{-2x} = \phi(x) + \psi(x)$
$4e^{-2x} = \phi'(x) - \psi'(x)$
integrate and rearrange to get:

$\phi(x) = e^{-2x}$ and $\psi(x) = 3e^{-2x}$

$$u = e^{-2(x+4t)} + 3e^{-2(x-4t)} \qquad x>4t$$

For the case of $0<x<4t$:
$u_x = \phi'(x+4t) + \psi'(x-4t)$ and $u$ are used in BC:

$\phi'(4t) + \psi'(-4t) - \phi(4t) -\psi(-4t) = e^{-8t}$

Then let $x = -4t$ and move the $\psi(x)$ to one side:

$\psi'(x) - \psi(x) = e^{2x} - \phi'(-x) + \phi(-x)$
We know two things: first the reason we're looking at $\psi$ instead of $\phi$ is that it has a chance of being negative, while the other does not, and thus, the second thing is that we can also plug in our previously used functions for $\phi$. The other nice thing is that you can rearrange to ODE:

$(e^{-x} \psi(x))' = e^{x} + (-2e^{x} + e^{x})$ integrate both sides:

$e^{-x} \psi(x) = C$
Thus:
$\psi(x) = Ce^{x}$

$u = e^{-2(x+4t)} + Ce^{x-4t}$

Then check continuity, plug in for $x = 4t$ for both u(x,t), and make sure they match up:

$u = e^{-16t} + C = e^{-16t} + 3$

Therefore $C= 3$, ensuring continuity.

EDIT: There is a chance I might have misplaced a minus sign, please let me know if I have made such an error, as I will aim to correct it promptly.

4
##### Final Exam / Re: FE-P3
« on: April 11, 2018, 05:35:17 PM »
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!

5
##### Quiz-B / Re: Quiz-B P2
« on: April 04, 2018, 12:08:01 PM »
Here's my initial attempt for the induction:

$< \partial f, \phi (x)> = <-f, \partial\phi (x)>$ for the general function $\phi(x)$. We can integrate by parts to get from LHS to get to RHS. This also holds for $\phi(x) = \delta(x)$. Then, for the n = 2 case, we can redefine a new function:

$<\partial^2 f, \phi(x)> = <-F, \partial^2\phi(x)>$ for the general function $\phi(x)$. We could also just redefine the functions in such a manner that we get the expression:

$<\partial F, \phi(x)> = <-f,\partial \phi(x)>$, it also holds for $\delta(x)$

Then generalizing to order  n = k, we can show it holds. Where F is the primitive for f. We can always reduce it to just a first order derivation instead of kth order.

6
##### Web Bonus Problems / Re: Web bonus problem--Week 10-11
« on: March 25, 2018, 09:12:44 PM »
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:

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

7
##### Term Test 2 / Re: TT2--P2N
« on: March 23, 2018, 08:38:42 PM »
As usual, let's setup our partial fourier transform, shifting from x basis to k basis, i.e.

$$\hat{f}(k) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ikx} f(x) dx$$

Thus equations (1,2,3) become

$$\hat{u_{yy}} - k^2 \hat{u} = 0 , \hat{u(k,0)} = 0 , (\hat{u_y}(k,1) + \alpha \hat{u}(k,1)) = \hat{g}(k)$$

Now, the fourier transform g(x) into $\hat{g}(k)$ is simply:

$\hat{g}(k) = \frac{1}{2\pi} \int_{-\infty}^{infty} e^{-ikx} g(x) dx = \frac{1}{2\pi} \int_{-1}^{1} e^{-ikx} dx = \frac{1}{2\pi} \frac{-e^{-ikx}}{ik} |_{-1}^{1} = \frac{1}{2\pi} \frac{e^{ik} - e^{-ik}}{ik} = \frac{1}{\pi k} sin(k)$

The solution to the differential equation from the fourier transform of (1) is:

$\hat{u} = A(k)e^{|k|y} +B(k)e^{-|k|y}$ ,  applying (2), we need: $-B(k) = A(k).$

$\hat{u} = A(k)e^{|k|y} -A(k)e^{-|k|y} = A(k)\sinh(|k|y)$

Finally, applying the condition (3):

$(A(k)|k| \cosh(|k|) + \alpha A(k)\\sinh(|k|) = \frac{sin(k)}{\pi k}$

So that: $A(k) = \frac{sin(k)}{\pi k} (|k|\cosh|k| + \alpha \sinh|k|)^{-1}$

Thus, our solution in x basis is given as:

$f(x) = \int_{-\infty}^{\infty} \hat{f(k)} e^{ikx} dk = \int_{-\infty}^{\infty} (\frac{sin(k)}{\pi k} (|k|\cosh|k| + \alpha \sinh|k|)^{-1})\sinh(|k|y)e^{ikx}dk$

8
##### Term Test 2 / Re: TT2--P3N
« on: March 23, 2018, 08:09:03 PM »
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

9
##### Term Test 2 / Re: TT2--P5
« on: March 23, 2018, 07:40:11 PM »
We first note that $\cos^2 x = (\frac{(e^{ix} + e^{-ix})}{2})^2 = \frac{2 + e^{2ix} + e^{-2ix}}{4}$

From there, we take the fourier transform:  $$\hat{f(}k) = \frac{1}{2\pi} \int_{-\infty }^{\infty } e^{-ikx} e^{-|x|} ( \frac{2 + e^{2ix} + e^{-2ix}}{4}) = \frac{1}{2\pi }\left ( \int_{0}^{\infty } \frac{2e^{-x(1+ki)} + e^{-x(1+ki - 2i)} + e^{-x(1+ki + 2i))}}{4} dx +\int_{-\infty }^{0} \frac{2e^{x(1- ki)} + e^{x(1- ki - 2i) }+ e^{x(1- ki + 2i))}}{4} dx\right )$$

We integrate that thing, and note that by evaluating at the bounds of $0,\pm \infty$, only the terms evaluated at 0 can be kept, netting us:

$$\hat{f(}k) = \frac{1}{2\pi}( \frac{1}{2(1+ki)} + \frac{1}{4(1+ki - 2i)} + \frac{1}{4(1+ki+2i)} + \frac{1}{2(1-ki)} + \frac{1}{4(1-ki-2i)} + \frac{1}{4(1-ki+2i)})$$

While this expression may be ugly, what is useful about it is the fact that these are fractions of complex numbers and their conjugates (noteably first and fourth terms, 2nd and 6th, 3rd and 5th) so we'd get the following fraction

$$\hat{f(}k) = \frac{1}{2\pi}( \frac{2}{4(1+ k^2)} + \frac{2}{16(1+ (k-2)^2)} + \frac{2}{16(1+ (k+2)^2)} )$$

Thus, we can plug this function into the IFT:

$$f(x) = \int_{-\infty}^{\infty} \hat{f}(k) e^{ikx} dk = \int_{-\infty}^{\infty} \frac{1}{4\pi}( \frac{1}{(1+ k^2)} + \frac{1}{4(1+ (k-2)^2)} + \frac{1}{4(1+ (k+2)^2)}) e^{ikx} dk$$

Leaving the function as a Fourier integral.

10
##### Quiz-5 / Quiz 5, T5102
« on: March 07, 2018, 06:28:31 PM »
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:

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

11
##### Term Test 1 / Re: P2 Night
« on: February 19, 2018, 11:55:31 AM »
With the initial conditions implying:

$g(x) = h(x) = 0$, and $c = 3$.
This means that D'Alembert's formula simplifies, from:

$$u(x,t) = \frac{1}{2} (g(x+ct)+g(x-ct)) + \frac{1}{2c} \int_{x-ct}^{x+ct} h(y)dy + \frac{1}{2c}\int_{0}^{t} \int_{x-c(t-t')}^{x+c(t-t')} f(x',t')dx'dt'$$

Into:

$$u(x,t) = \frac{1}{6}\int_{0}^{t} \int_{x-3(t-t')}^{x+3(t-t')} f(x',t')dx'dt'$$

As suggested, rearranging for the order of integration would be convenient here. For simplicity, I also broke up the characteristic triangle into two smaller triangles.

For the first one:

$0 < t' < x \frac{x'-x+3t}{3}$ and $x-3t < x' < x$

And the second:

$0 < t' < x \frac{-x'+x+3t}{3}$ and $x < x' < x+3t$

This breaks our integral up into the following:

$$(1) = \frac{1}{6} \int_{x-3t}^{x} \int_{0}^{\frac{x' - x + 3t}{3}} 18e^{-x'^2}dt'dx'$$
$$(2) = \frac{1}{6} \int_{x}^{x+3t} \int_{0}^{\frac{-x' + x + 3t}{3}} 18e^{-x'^2}dt'dx'$$
Where $u(x,t) = (1)+(2)$

Starting with (1), we first integrate to get:

$$(1) = \frac{1}{6} \int_{x-3t}^{x} \frac{x' - x +3t}{3} 18e^{-x'^2} dx'$$
$$(1) = 3 \int_{x-3t}^{x} (3t-x)e^{-x'^2}dx' + 3 \int_{x-3t}^{x} x'e^{-x'^2} dx'$$

The first integral can be resolved to error functions, the second requires a simple substitution to solve:
Let

$u = x^2$ then $du = 2xdx$. Thus, for second term $\int_{(x-3t)^2}^{x^2} \frac{3}{2} e^-u du = \frac{3}{2} ( e^{-(x-3t)^2} - e^{-(x)^2})$

Putting it together, we get that :

$$(1) = 3\sqrt{\pi} (3t-x) (erf(x) - erf(x-3t)) + \frac{3}{2} ( e^{-(x-3t)^2} - e^{-(x)^2})$$

Likewise, for (2) we should get:

$$(2) = 3\sqrt{\pi} (3t+x) (erf(x+3t) - erf(x)) - \frac{3}{2} ( e^{-(x)^2} - e^{-(x+3t)^2})$$

Then

$$u(x,t) = 3 \sqrt{\pi} (3t(erf(x+3t) - erf(x-3t)) - 2xerf(x) + x(erf(x+3t)+erf(x-3t)) - 3e^{-x^2} + \frac{3}{2} (e^{-(x-3t)^2} + e^{-(x-3t)^2}$$

12
##### Term Test 1 / Re: P1 Night
« on: February 16, 2018, 03:51:56 PM »
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:

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

13
##### Term Test 1 / Re: P4 Night
« on: February 16, 2018, 03:24:59 PM »
First take $\frac{dE}{dt}$ of (4):

We get $$\int_{0}^{L} (u_t u_{tt} + c^2 u_{xx}u_{xxt} + a u u_t ) dx$$

Now, integrate the middle term by parts, twice. The nice thing is that due to those boundary conditions, we can simply write:

$$\int_{0}^{L} (u_t u_{tt} - c^2 u_{xxx}u_{xt} + a u u_t ) dx + c^2u_{xx}u_{xt}|_{0}^{L}$$
The last term evaluates to 0 at both 0 and L, so redoing this integration by parts gives us:

$$\int_{0}^{L} (u_t u_{tt} + c^2 u_{xxxx}u_{t} + a u u_t ) dx + c^2u_{xxx}u_t|_{0}^{L}$$
Which again allows for one of the terms $u_{xx}$ and $u_{xxx}$ to be evaluated as 0 at 0 and L.

$$\int_{0}^{L} u_t ( u_{tt} + c^2 u_{xxxx} + a u ) dx$$

The factored term inside is clearly the PDE we set to be 0 (eqn 1).

Therefore our integrand is 0, and thus

$$\frac{dE}{dt} = 0$$

Therefore energy is conserved.

14
##### Term Test 1 / Re: P5
« on: February 16, 2018, 03:15:35 PM »
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.

15
##### Quiz-3 / Re: Q3-T0101
« on: February 11, 2018, 06:34:50 PM »
I got something similar, but quite the same as Jingxan:

$$u = \phi (x+ct) , x> ct$$

On the region $0<x<ct$ , we get from the boundary condition:

$$0 = \Theta '(ct) + \Psi' (-ct) + \alpha (\Theta(ct) + \Psi(-ct))$$

Apply that $x = -ct$, and this prompts for me to rearrange and treat each side as an ODE:

$$\Psi'(x) + \alpha \Psi(x) = -(\Theta'(-x) + \alpha \Theta(-x))$$
$$(e^{\alpha x} \Psi(x) ) ' = -(\Theta'(-x) + \alpha \Theta(-x))$$
And, critically:
$$e^{\alpha x} \Psi(x)) = \Theta(-x) + \alpha \int_{0}^{-x} \Theta(x')dx' + c$$

$$\Psi(x) = e^{-\alpha x}(\Theta(-x) + \alpha \int_{0}^{-x} \Theta(x')dx' + c$$

Leading to: (note $\Theta = \phi$ in this problem)

$$u (x,t) = \phi(x+ct) + e^{\alpha (ct-x)}(\phi(ct-x) + \alpha \int_{0}^{ct-x} \Theta(x')dx' + k )$$

Examining part b: $\phi(x) = e^{ikx}$ :

$$u(x,t) = e^{ik(x+ct)} + e^{\alpha( ct - x)}( e^{ik(ct-x)} + \alpha \int_{0}^{ct-x} e^{ikx'}dx' + c)$$

$$u(x,t) = e^{ik(x+ct)} + e^{\alpha( ct - x)}( e^{ik(ct-x)} + \alpha (\frac{e^{ik(ct-x)} - 1}{ik} + c)$$

Which can be expanded into sines and cosines:

$$u(x,t) = cosk(x+ct) + i sink(x+ct) + e^{\alpha} cos(k(ct-x)) + \alpha e^{\alpha}(\frac{cosk(ct-x) + isink(ct-x) - c}{ik})$$

A few notes:

I share the same concerns as George about the $-2\alpha\phi(-s)$, and I'm not quite sure if my above solution is correct. What I do know about part b is that we are probably expected to take the mess of sines and cosines we get, and rearrange them into some form of $sin( a \pm b)$ or $cos(a \pm b)$ or we could maybe break it into $Re$ and $Im$ components?

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