Author Topic: Problem 2  (Read 6063 times)

Calvin Arnott

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Problem 2
« on: October 04, 2012, 03:26:12 PM »
It seems to me that in question 2 in homework 3, the letter "v" is used to refer to both a constant, and a function u(x,t) = v(x - v t, t). Under this interpretation everything works out nicely, and I assume this to be the case.

Is this the correct? Or is v a function which recursively calls itself? If not, it might be worth changing the question to have a different letter to represent the constant or function.

Victor Ivrii

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Re: Problem 2
« Reply #1 on: October 04, 2012, 04:10:57 PM »
It seems to me that in question 2 in homework 3, the letter "v" is used to refer to both a constant, and a function u(x,t) = v(x - v t, t). Under this interpretation everything works out nicely, and I assume this to be the case.

Is this the correct? Or is v a function which recursively calls itself? If not, it might be worth changing the question to have a different letter to represent the constant or function.

Good catch--updated!

Levon Avanesyan

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Re: Problem 2
« Reply #2 on: October 06, 2012, 02:37:19 PM »
In this problem it is stated that we should use formulas (1)-(2). But shouldn't we have an initial condition in order to apply these formulas.

Victor Ivrii

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Re: Problem 2
« Reply #3 on: October 06, 2012, 03:16:06 PM »
In this problem it is stated that we should use formulas (1)-(2). But shouldn't we have an initial condition in order to apply these formulas.

Sure we need: so add $u(x,0)=g(x)$.

Peishan Wang

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Re: Problem 2
« Reply #4 on: October 08, 2012, 06:04:48 AM »
Professor for part (b), is there any restriction on when we can use method of reflection (continuation) to solve IBVP? Thanks!

Victor Ivrii

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Re: Problem 2
« Reply #5 on: October 08, 2012, 08:08:30 AM »
Professor for part (b), is there any restriction on when we can use method of reflection (continuation) to solve IBVP? Thanks!

It was discussed several times on forum and in the lecture 10: http://www.math.toronto.edu/courses/apm346h1/20129/L10.html




Peishan Wang

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Re: Problem 2
« Reply #6 on: October 08, 2012, 09:39:08 AM »
Yeah I've read the lecture notes twice.....  actually my question is, can we use method of reflection (continuation) only with a boundary condition at x=0? From what we've learned, we use this method only when we have a Dirichlet or Neumann boundary condition (evaluated at x=0). If I instead have a boundary condition evaluated at, say x=t, is there any technique that allows me to take advantage of the method of continuation? I don't know how I can make the boundary condition automatically satisfied in this case..... Thanks.
« Last Edit: October 08, 2012, 09:57:04 AM by Peishan Wang »

Victor Ivrii

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Re: Problem 2
« Reply #7 on: October 08, 2012, 09:55:12 AM »
Yeah I've read the lecture notes twice.....  actually my questions is, can we use method of reflection (continuation) only with a boundary condition at x=0? From what we've learned, we use this method only when we have a Dirichlet or Neumann boundary condition (evaluated at x=0). If I instead have a boundary condition evaluated at, say x=t, is there any technique that allows me to take advantage of the method of continuation? I don't know how I can make the boundary condition automatically satisfied in this case..... Thanks.

You may try to reduce boundary at $x=a$ to $x=0$ by shift $x_{\text{new}}=x-a$. Boundary at $x=v t$ seems to be a non-starter.

Hint for those who really want to apply this method to $u_{t}+v u_{x}-k u_{xx}=0$. Instead of change of variables in this equation leading us to normal heat equation (without convection term) try
$$
u = e^{\alpha t + \beta x} V(x,t)
$$
which does not break the boundary at $x=0$; one can find constants $\alpha$ and $\beta$ s.t. $V$ should satisfy normal heat equation.


Zarak Mahmud

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Re: Problem 2
« Reply #8 on: October 10, 2012, 09:34:11 PM »
Part (a):

$$\begin{equation*}
u(x,t) = U(x - vt, t)\\
Let \xi = x - vt, \eta = t \\
 u_x = U_x = \frac{\partial U}{\partial \xi}\frac{\partial \xi}{\partial x} + \frac{\partial U}{\partial \eta}\frac{\partial \eta}{\partial x} = U_x\\
u_{xx} = U_{\xi\xi}\\
u_t = U_t = \frac{\partial U}{\partial \xi}\frac{\partial \xi}{\partial t} + \frac{\partial U}{\partial \eta}\frac{\partial \eta}{\partial t} = -vU_{\xi} + U_{\eta}\\

 -vU_{\xi} + U_{\eta} + vU_{\xi} = kU_{\xi\xi} \\
 U_{\eta} = kU_{\xi\xi}\\


U(\xi,\eta) = \int_{-\infty}^{\infty}G(\xi,y,\eta)g(y)dy\\
u(x,t) = \int_{-\infty}^{\infty}G(x - vt,y,t)g(y)dy \\
= \int_{-\infty}^{\infty} \frac{1}{2 \sqrt{k \pi t}} e^{\frac{-(x-vt-y)^2}{4kt}}g(y)dy
\end{equation*}$$

Part (b):

Let $$
\begin{equation*}
u(x,t) = e^{\alpha t + \beta x}V(x,t)\\
u_t = \alpha e^{\alpha t + \beta x}V + e^{\alpha t + \beta x}V_t\\
u_x = \beta e^{\alpha t + \beta x} + e^{\alpha t + \beta x}V_t\\
u_{xx} = \beta^2 e^{\alpha t + \beta x}V + 2\beta e^{\alpha t + \beta x}V_x + e^{\alpha t + \beta x}V_{xx}\\
\end{equation*}$$

Now plug these into the given heat equation.

$$
\begin{equation*}
\alpha V + V_t + v\beta V + vV_x = k\beta^2 V + 2k\beta V_x + kV_{xx} \end{equation*}$$
set $ \alpha + v\beta = k\beta^2$ and $ v = 2k\beta$ then,

$$\begin{equation*}
\beta = \frac{v}{2k}, \alpha = \frac{-v^2}{4k} \\

u(x,t) = e^{\alpha t + \beta x} \int_{0}^{\infty} G_D(x,y,t)g(y)dy\end{equation*} \\$$
This works because if $u(0,t) = 0,$ then $ V(0,t) = 0.$
« Last Edit: October 10, 2012, 11:02:36 PM by Zarak Mahmud »

Qitan Cui

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Re: Problem 2
« Reply #9 on: October 10, 2012, 11:07:51 PM »
Here is my solution to this problem

I used the same approach as Zarak did, but in part (d) I think we would have to consider Dirichlet and Neumann boundary conditions separately. For Dirichlet, u(t,0)=> V(t,0) and we can use method of continuation to solve the V(x,t) first and then get u(x,t).

But for Neumann boundary condition, ux(t,0)=0  => nV(t,0) + Vx(t,0)=0 which does not necessarily mean that V(t,0)=0 or Vx(t,0)=0. In this case the boundary conditions are not automatically satisfied and we might not be able to use method of continuation.

Please let me know if there's anything wrong with this solution. Thanks in advance!
« Last Edit: October 10, 2012, 11:25:20 PM by Qitan Cui »

Victor Ivrii

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Re: Problem 2
« Reply #10 on: October 11, 2012, 05:11:34 AM »
This problem has many layers. First:

(i) We cannot use continuation method directly to original equation because convection term contains odd derivative by $x$
(ii) Change of variables removes this derivative but injects a moving boundary. This change is natural as this equation was obtained from the ordinary heat equation by the opposite change (so we consider heat exchange but in the constant flow of air) which makes equation more complicated--but allows us to consider boundaries as stationary.
(iii) However the substitution $u(x,t)= e^{\alpha x +\beta t}U(x,t)$ resolves this issue.

Sure we should consider both D. and N. and
(i) with D. everything is fine and
(ii) Qui is completely right noting that it is not the case with Neumann as Neumann becomes Robin $U_x + \alpha U=0$ as $x=0$.
(iii) However if original condition was Robin $u_x + \gamma u=0$ as $x=0$, then we get Robin again $U_x +(\gamma+ \alpha) U=0$ as $x=0$ and for $\gamma=-\alpha$ we get Neumann and therefore for some very special Robin for $u$ we can apply method of continuation.

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