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Messages - Emily Deibert

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Website errors / Wrong lecture room listed on website
« on: October 18, 2015, 05:13:41 PM »
Hi Professor,

I just noticed that on the website you have our lecture room listed as SF 1101, however our lecture is in Sid Smith! The room should be SS 2110.

Textbook errors / Re: Section2.3 equation 2
« on: October 18, 2015, 04:50:37 PM »
Also in the following line for equation2
 We have "w=(∂t−c∂x)u=ut−cux"
 Shouldn't it be "w=(∂t−c∂x)u=wt - wux"

But why would the w come into the equation here? I think we are saying that \begin{equation}
w = (\partial_t - c\partial_x)u = \partial_tu - c\partial_xu \end{equation}
As in, we are expanding what is in the brackets so that it acts on u, and thus this is what w is equal to. Maybe I am misunderstanding?

Edit: I agree with the original post though, I think a u is missing.

HA5 / Re: HA5-P3
« on: October 18, 2015, 04:46:19 PM »
Oh I see, this makes sense. I think I misunderstood what the question was asking.

Textbook errors / Re: 3.1 equation 12
« on: October 17, 2015, 11:01:31 PM »
Shouldn't there also be a $\frac{1}{2}$ as part of $U(x,t)$? Or else where does the extra term of $\frac{1}{2}$ come from?

HA5 / Re: HA5-P3
« on: October 17, 2015, 08:17:51 PM »
I'll add my typed solutions.

(a) The ordinary heat equation is: \begin{equation}
u_t = ku_{xx} \end{equation}

Now consider $U(x,t) = u(x+ct,t)$. The partial derivatives needed for the heat equation are given by: \begin{equation}
U_t = cu_x + u_t \\
U_x = u_x \\
U_{xx} = u_{xx} \end{cases} \end{equation}

Now let's plug these into the heat equation: \begin{equation}
U_t = kU_{xx} \longrightarrow (u_t + cu_x) = k(u_{xx}) \longrightarrow u_t + cu_x = ku_{xx} \end{equation}

Therefore the heat equation with a convection term is obtained from the ordinary heat equation with a change of variables.

(b) Now we use the change of variabes $u(x,t) = U(x-ct,t)$ (Note: Thank you Rong Wei for pointing out the fact that this should be c here, as I was unable to solve the problem otherwise!). Now the partial derivatives are given by: \begin{equation} \begin{cases}
u_t = -cU_x + U_t \\
u_x = U_x \\
u_{xx} = U_{xx} \end{cases} \end{equation}

Let's plug this into the heat equation with a convective term: \begin{equation}
u_t + cu_x = ku_{xx} \longrightarrow (-cU_x + U_t) + c(U_x) = k(U_{xx}) \longrightarrow -cU_x + cU_x + U_t = kU_{xx} \longrightarrow U_t = kU_{xx}
So with this change of variables, the equation reduces to the familiar heat equation.
We can then use the usual formula to arrive at the solution: \begin{equation}
u(x,t) = \int_0^{\infty}G(x-ct,y,t)g(y)dy =  \frac{1}{2\sqrt{\pi{}kt}}\int_0^{\infty}e^{-(x-ct-y)^2/(4kt)}g(y)dy \end{equation}

(c) I don't think we can use the method of continuation directly to solve IVBP with Dirichlet or Neumann boundary conditions as $x>0$ for the heat equation with a convection term on ${x>0, t>0}$. This is because we used a change of variables to define the problem.

(d)We have \begin{equation}
u(x,t) = v(x,t)e^{\alpha{}x + \beta{}t} \end{equation}

The partial derivatives are given by: \begin{equation} \begin{cases}
u_t = v_te^{\alpha{}x + \beta{}t} + \beta{}ve^{\alpha{}x + \beta{}t} \\
u_x = v_xe^{\alpha{}x + \beta{}t} + \alpha{}ve^{\alpha{}x + \beta{}t} \\
u_{xx} = v_{xx}e^{\alpha{}x + \beta{}t} + 2\alpha{}v_xe^{\alpha{}x + \beta{}t} + \alpha{}^2ve^{\alpha{}x + \beta{}t} \end{cases} \end{equation}

Let's plug them in to the heat equation with a convection term. \begin{equation}
v_te^{\alpha{}x + \beta{}t} + \beta{}ve^{\alpha{}x + \beta{}t} + cv_xe^{\alpha{}x + \beta{}t} + c\alpha{}ve^{\alpha{}x + \beta{}t} = ke^{\alpha{}x + \beta{}t}(v_{xx} + 2\alpha{}v_x + \alpha{}^2v) \end{equation}

Gathering like terms, we arrive at: \begin{equation}
v_t + v(\beta{} + c\alpha{} - k\alpha{}^2) + v_x(c-2k\alpha{}) = kv_{xx} \end{equation}

For this to reduce to the heat equation, we want the coefficients in front of $v$ and $v_x$ to be zero. So: \begin{equation} \begin{cases}
\beta{} + c\alpha{} - k\alpha{}^2 \\
c-2k\alpha{} \end{cases} \end{equation}

Solving first for $\alpha{}$, we get that $\alpha{} = \frac{c}{2k}$. We can then plug this into the equation for $\beta{}$ and solve to get that $\beta{} = \frac{-c^2}{4k}$.

(e) I will work on this part in a little while. So far I think I have gotten all of the same solutions as Rong Wei. Added my solution below.
For the case of the half-line and Dirichlet boundary condition, we will have the solution: \begin{equation}
u(x,t) = \frac{e^{\alpha{}x + \beta{}t}}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{-(x-y)^2/4kt} - e^{-(x+y)^2/4kt}]g(y)dy  \end{equation}
In the case of Neumann boundary conditions, we cannot use a similar method.

HA5 / HA5-P7
« on: October 17, 2015, 06:41:15 PM »
I will update my answers as I finish working through them. All of my answers are now uploaded.

(a) We consider the energy as given in the problem: \begin{equation}
E(t) = \int_{J} u^2(x,t)dx

Now, take the time derivative to yield: \begin{equation}
\frac{\partial}{\partial t}E(t) = \int_{J} \frac{\partial}{\partial t}u^2(x,t)dx = \int_{J} 2uu_tdx \end{equation}

Replacing $u_t$ with $ku_{xx}$ as per the heat equation, we get: \begin{equation}
\int_J 2kuu_{xx}dx \end{equation}

Now, using integration by parts, we will get: \begin{equation}
2k\Big[uu_x|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} u_x^2dx\Big] \end{equation}

Here I assume that the function is fast-decaying, so the first term goes to 0 (professor, is this a valid assumption?) This will leave me with just:  \begin{equation}
-2k\int_{-\infty}^{\infty} u_x^2dx  \label{eq:integral}\end{equation}

Now consider \ref{eq:integral}. Since the integrand is positive (since it is $u_x^2$) and k is positive (by the definition of the heat equation we assume this), then this equation is $\leq 0$.

If $u(x,t) = $ const., then the integrand will be zero (indeed, $\frac{d}{dx}C = 0$).

In all other cases, this is < 0. Therefore $E(t)$ does not increase; further, it decreases unless $u(x,t) = $ const.

(b) This part proceeds in the same way up to the integration by parts (except with J defined as the range from 0 to L rather than negative to positive infinity), so I will not type all of those steps again. Just recall that: \begin{equation}
\frac{\partial}{\partial t}E(t) = 2k\Big[uu_x|_0^L - \int_0^Lu_x^2dx\Big] \label{eq:newint} \end{equation}

We consider the cases of both Dirichlet and Neumann boundary conditions. First, let's think about Dirichlet. This condition says that: \begin{equation}
u|_{x=0} = u|_{x=L} = 0 \end{equation}
Therefore the first term in \ref{eq:newint} goes to zero by the Dirichlet boundary condition on $u$.

Now let's consider the Neumann boundary condition. It says that: \begin{equation}
u_x|_{x=0} = u_x|_{x=L} = 0. \end{equation}
Again, the first term in \ref{eq:newint} will go to zero, this time by the Neumann boundary condition on $u_x$.
In both cases, we see that the end result is:  \begin{equation}
\frac{\partial}{\partial t}E(t) = -2k\int_0^Lu_x^2dx \end{equation}
We proceed as in part (a), showing that $E(t)$ decreases unless $U(x,t) = $ const because the integrand and the constant k are both positive.

(c) This problem is identical to (a) and (b) up to the integration by parts (where J is defined as the region from 0 to L, as in part (b)), so I won't type up the first few steps again. We arrive at: \begin{equation}
\frac{\partial}{\partial t}E(t) = 2k\Big[uu_x|_0^L - \int_0^Lu_x^2dx\Big]

This time, we have Robin boundary conditions. They are: \begin{equation}
u_x(0,t) - a_0u(0,t) = 0 \\
u_x(L,t) + a_Lu(L,t) = 0

The first boundary condition says that at $x=0$, $u=\frac{u_x}{a_0}$. The second says that at $x=L$, $u=\frac{u_x}{-a_L}$. So now let's consider the first term in the integral in \ref{eq:c}: \begin{equation}
uu_x|_0^L = u_x(L,t)\frac{u_x(L,t)}{-a_L} - u_x(0,t)\frac{u_x(0,t)}{a_0} = \frac{u_x^2}{-a_L} - \frac{u_x^2}{a_0} = -\Big(\frac{u_x^2(L,t)}{a_L} + \frac{u_x^2(0,t)}{a_0}\Big)

So, now let's put this back into \ref{eq:c} to yield: \begin{equation}
\frac{\partial}{\partial t}E(t) = -2k\Big[\Big(\frac{u_x^2(L,t)}{a_L} + \frac{u_x^2(0,t)}{a_0}\Big) + \int_0^Lu_x^2dx\Big]

Since all functions and the integrand are to the power of 2, they are all positive; all constants ($a_L$, $a_0$, and k) are also defined to be positive. This implies that overall, \begin{equation}
-2k\Big[\Big(\frac{u_x^2(L,t)}{a_L} + \frac{u_x^2(0,t)}{a_0}\Big) + \int_0^Lu_x^2dx\Big] \leq 0

Therefore the energy is decreasing or zero, and in fact the endpoints contribute to this decrease. Note that in this case the Robin boundary condition prevents a constant solution from being possible, since for $C\neq0$: \begin{equation}
\frac{d}{dx}C -a_0C = 0 - a_0C  = -a_0C\neq 0 \end{equation}
(And likewise for the other condition).
The only case where a constant solution is valid is the trivial solution $u(x,t) = 0$. In this case only will the equality to zero occur.
Therefore the endpoints contribute to the decrease of $E(t)$.

Textbook errors / Missing reference in HA5 (Problems to 3.2)
« on: October 17, 2015, 06:08:03 PM »
Hi Professor, I noticed that in Remark 1 in HA5 question 7, you have a link to an equation written as (???).

Fixed. It is not a missing but broken reference. In LaTeX if \ref {mylabel} is written (without space) (and \ref has some variants like \pageref, \footref and some package-specific) but \label{mylabel} is missing (or misspelled) then \ref{mylabel}  appears.  MathJax has only \ref

PS On our forum \ref{x} appears as an emoticon ??? (one needs to be careful with these things on forums, blogs, etc)

Oops, sorry about that! -Emily

HA5 / Re: HA5-P1
« on: October 17, 2015, 06:00:50 PM »
@ Rong Wei, I believe we both got the same solution, just that you wrote yours in terms of $G_0$ and I wrote mine in terms of the exponential definition of G.

HA5 / Re: HA5-P1
« on: October 17, 2015, 05:48:59 PM »
(a) Here is my typed solution:

We have the Dirichlet problem \begin{equation}
u_t = ku_{xx} & x>0, t>0 \\
u|_{t=0} = g(x) \\
u|_{x=0} = 0

By method of continuation, we extend the initial function to the full line (recalling the odd extension for Dirichlet problems): \begin{equation}
g_{extended}(x) = \begin{cases} g(x) & x>0 \\
-g(-x) & x<0 \\
0 & x=0

So for the entire line (i.e. $-\infty<x<\infty$), we arrive at a solution with the formula as given in the textbook; call it w. The solution to the given problem \ref{eq:problem} will just be this solution in the region of x>0. \begin{equation}
w(x,t) = \int_{-\infty}^{\infty} G(x,y,t)g_{extended}(y)dy
So using the definition of $g_{extended}$ as given in \ref{eq:gext}, we can rewrite this as: \begin{equation}
w(x,t) = \int_{0}^{\infty} G(x,y,t)g(y)dy - \int_{-\infty}^{0} G(x,y,t)g(-y)dy
In the second equation above, let's make a change of variables from -y to y. Then this yields: \begin{equation}
w(x,t) = \int_0^{\infty}[G(x,y,t) - G(x, -y, t)]g(y)dy \end{equation}
This is our solution. Written in terms of the definition of G, we arrive at the final solution to \ref{eq:problem}: \begin{equation}
u(x,t) = \frac{1}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{-(x-y)^2/(4kt)} - e^{-(x+y)^2/(4kt)}]g(y)dy \end{equation}

(b) Part (b) is very similar, only here we have a Neumann problem: \begin{equation}
u_t = ku_{xx} & x>0, t>0 \\
u|{t=0} = g(x) \\
u_x|{x=0} = 0

Recall that for Neumann problems, the continuation is even. So we write the extension of the initial function to the entire line: \begin{equation}
g_{extended}(x) = \begin{cases} g(x) & x \geq 0 \\
g(-x) & x \leq 0 \end{cases}

Proceeding similarly as in part (a), the solution for the whole line will be: \begin{equation}
w(x,t) = \int_{-\infty}^{\infty} G(x,y,t)g_{extended}(y)dy
The solution to \ref{eq:neumann} will just be this solution in the region x>0.
So let's write this in terms of the definition in \ref{eq:geven}: \begin{equation}
w(x,t) = \int_{0}^{\infty} G(x,y,t)g(y)dy + \int_{-\infty}^{0} G(x,y,t)g(-y)dy
Let's again make the change of variables, from -y to y. We arrive at: \begin{equation}
w(x,t) = \int_{0}^{\infty} [G(x,y,t) + G(x,-y,t)]g(y)dy
Using the definition of G as before, we arrive at the final solution to \ref{eq:neumann}: \begin{equation}
u(x,t) = \frac{1}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{-(x-y)^2/(4kt)} + e^{-(x+y)^2/(4kt)}]g(y)dy \end{equation}

HA4 / Re: HA4-P3
« on: October 15, 2015, 05:45:27 PM »
From my understanding v is just some constant that could have any value. We consider what conditions will be necessary for the various cases of v in relation to c (as mentioned in the problem).

HA4 / HA4-P1
« on: October 09, 2015, 05:07:44 PM »
Problem 1:

(a) Here we have a Dirichlet boundary condition. In the domain ${t>0, x \geqslant ct}$, the solution is given by the D'Alembert formula:
u(x,t) = \frac{1}{2}[g(x+ct) + g(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} h(y)dy
So plugging in the initial conditions as specified in the problem, we arrive at the solution:
u(x,t) = \phi(x+ct)
In the domain ${0<x<ct}$, the solution is given by:
u(x,t) = \frac{1}{2}g(x+ct) + \frac{1}{2c} \int_{0}^{x+ct} h(y)dy + p\bigg(t - \frac{x}{c}\bigg) - \frac{1}{2}g(ct-x) - \frac{1}{2c} \int_0^{ct-x} h(y)dy
So plugging in the initial and boundary conditions as specified in the problem, we arrive at the solution:
u(x,t) = \phi (x+ct) + \chi\bigg(t-\frac{x}{c}\bigg) - \phi(ct-x)
So the solution to (a) is
u(x,t) =
\phi(x+ct) & \text{for } \{t>0, x \geqslant ct\}\\
\phi (x+ct) + \chi\bigg(t-\frac{x}{c}\bigg) - \phi(ct-x) & \text{for } \{0<x<ct\}

(b) In this case we have a Neumann boundary condition. The solution for the domain ${t>0, x \geqslant ct}$ is the same as in part (a), namely:
u(x,t) = \phi(x+ct)
For the domain ${0<x<ct}$, the solution is given by:
u(x,t) = \frac{1}{2}g(x+ct) + \frac{1}{2c} \int_0^{x+ct} h(y)dy - c \int_0^{t-\frac{x}{c}} q(t')dt' + \frac{1}{2}g(ct-x) + \frac{1}{2c} \int_0^{ct-x} h(y)dy
Plugging in our initial and boundary conditions as specified in the problem, we arrive at the solution:
u(x,t) = \phi (x+ct) - \phi (0) - cX\bigg(t-\frac{x}{c}\bigg) + cX(0) + \phi(ct-x)
\int\chi(t')dt' = X(t')
So the solution to (b) is
u(x,t) =
\phi(x+ct) & \text{for } \{t>0, x \geqslant ct\}\\
\phi (x+ct) - \phi (0) - cX\bigg(t-\frac{x}{c}\bigg) + cX(0) + \phi(ct-x) & \text{for } \{0<x<ct\}

Quiz 2 / Solution to Quiz 2
« on: October 08, 2015, 10:57:27 PM »
Here is my solution. The process is the same as HA3-Problem2 ( so I will not post that. Please let me know if you notice any errors or typos.

Professor, please excuse my ugly graph---I do not know how to make one as nice as you did in Problem 2! ( I also don't know how to put the image in the post as you did, so I will attach it here.

Textbook errors / Broken link in web bonus problem week 4 #6
« on: October 08, 2015, 03:37:58 PM »

HA3 / Re: HA3-P2
« on: October 08, 2015, 01:41:44 PM »
Professor, I am a little confused about the different regions. Can you please explain this?

Textbook errors / Typo in HA3 P3?
« on: October 03, 2015, 05:54:34 PM »
I asked this in the board for HA3 P3 as well but I am cross-posting here.

Just wondering, is there a typo in the original problem? The equation given is: \begin{equation}
Au_{tt} + 2Bu_{tx} + Cu_{tt} \end{equation}
But I think the last term should be with respect to x: \begin{equation}
Au_{tt} + 2Bu_{tx} + Cu_{xx}
Indeed, corrected

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