Author Topic: Problem 1  (Read 12227 times)

Kun Guo

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Problem 1
« on: November 10, 2012, 03:51:49 PM »
why in problem 1 we are looking for 'solutions' instead of a single solution?

Ziting Zhou

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Re: Problem 1
« Reply #1 on: November 10, 2012, 03:53:38 PM »
why in problem 1 we are looking for 'solutions' instead of a single solution?

Haha, I have the same question. I can only find one solution. What do you guys think?

Yishen Song

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Re: Problem 1
« Reply #2 on: November 10, 2012, 04:21:25 PM »
Maybe due to different constants.

Ziting Zhou

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Re: Problem 1
« Reply #3 on: November 10, 2012, 04:22:32 PM »
Maybe due to different constants.

I think you are right.

Danny Dinh

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Re: Problem 1
« Reply #4 on: November 17, 2012, 08:35:21 PM »
Are we supposed to use spherical coordinates for this question? If so, should r be replaced with rho?

Ian Kivlichan

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Re: Problem 1
« Reply #5 on: November 19, 2012, 09:20:53 PM »
Are we supposed to use spherical coordinates for this question? If so, should r be replaced with rho?
You can call the variables whatever you feel like - it makes no difference.

Fanxun Zeng

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Re: Problem 1
« Reply #6 on: November 19, 2012, 09:29:56 PM »
Problem 1 a solution attached

Fanxun Zeng

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Re: Problem 1
« Reply #7 on: November 19, 2012, 09:30:30 PM »
Problem 1 b solution attached

Ian Kivlichan

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Re: Problem 1
« Reply #8 on: November 19, 2012, 09:30:53 PM »
Hopeful solutions attached! :)

(Essentially the same as Fanxun's, but with exponential instead of sines and cosines.. the difference is trivial.)
« Last Edit: November 19, 2012, 09:32:30 PM by Ian Kivlichan »

Chen Ge Qu

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Re: Problem 1
« Reply #9 on: November 19, 2012, 09:32:05 PM »
Both parts attached

Calvin Arnott

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Re: Problem 1
« Reply #10 on: November 19, 2012, 09:35:54 PM »
Problem 1

Part a Let $ k > 0 $. Find the solutions that depend only on $r$ of the equation

$$ \Delta u\left(x,y,z\right) := u_{xx} + u_{yy} + u_{zz} = k^2 u    $$

Answer:
As previously derived, transforming $u\left(x,y,z\right)$ into spherical coordinates $u\left(r,\theta,\phi\right)$ yields the PDE:

$$ \Delta u\left(r,\theta,\phi\right) := u_{rr} + \frac{2}{r} u_{r} + \frac{1}{r^2}\left(u_{\theta\theta} +\cot\left(\theta\right)u_\theta + \frac{1}{\sin\left(\theta\right)^2}u_{\phi\phi}\right) = k^2 u $$

Since we seek only $u\left(r,\theta,\phi\right)$ which depends $r$, we have that $u\left(r\right)$ is constant with respect to both of $\{\theta,\phi\}$. Then $u_{\theta\theta} = u_{\theta} = u_{\phi\phi} = 0$, and our equation simplifies to an ODE:

$$ \Delta u\left(r\right) := u_{rr} + \frac{2}{r} u_{r} = k^2 u $$

Substituting in $ u\left(r\right) = \frac{v\left(r\right)}{r} \implies u_{r} = \frac{v_r}{r} - \frac{v}{r^2} $, $u_{rr} = \frac{v_{rr}}{r} - \frac{v_{r}}{r^2} - \frac{v_{r}}{r^2}  + \frac{2 v}{r^3} $ yields:

$$  \Delta u := \left(\frac{v_{rr}}{r} - \frac{2 v_{r}}{r^2} + \frac{2 v}{r^3}\right) + \frac{2}{r} \left( \frac{v_r}{r} - \frac{v}{r^2} \right) = k^2  \frac{v}{r}  $$

Multiplying through by $r$ and simplifying terms then gives us a nice expression:

$$ v_{rr} = k^2 v $$

Since we have that $ k > 0 $, $k^2 > 0$, our ODE has the solution, with $ \{A,B,C,D\} \in \mathbb{R} $:

$$ v\left(r\right) = A e^{k r} + B e^{-k r} = C \cosh\left(k r\right) + D \sinh\left(k r\right) $$

$$ \implies u\left(r\right) = \frac{1}{r} v\left(r\right) = \frac{1}{r}\left(A e^{k r} + B e^{-k r}\right) =\frac{1}{r} \left(C \cosh\left(k r\right) + D \sinh\left(k r\right)\right) \phantom{\ } \blacksquare$$



Part b. Let $ k > 0 $. Find the solutions that depend only on $r$ of the equation

$$ \Delta u := u_{xx} + u_{yy} + u_{zz} = -k^2 u    $$


Answer:
Following the derivation of part a., we substitute in spherical coordinates: $u\left(x,y,z\right) \rightarrow u\left(r,\theta,\phi\right)$, assume a form of $u\left(r,\theta,\phi\right)$ which depends only on $r$, $u\left(r\right)$. Then make the substitution $ u\left(r\right) = \frac{v\left(r\right)}{r} $ and simplify the resulting ODE. Because the left hand side of our equation is unchanged between part a. and b., this derivation yields identical equations on the left hand side. The difference in the two problems appears only in the right hand side of the expression, where $ \Delta u = -k^2 u $ in place of $ \Delta u = k^2 u $. After simplification this yields the ODE:

$$ v_{rr} = - k^2 v $$

Which, as $k > 0$, $-k^2 < 0$ has a solution in the form, for some $ \{A,B,C,D\} \in \mathbb{R} $:

$$ v\left(r\right) = A e^{i k r} + B e^{-i k r} = C \cos\left(k r\right) + D \sin\left(k r\right) $$

$$ \implies u\left(r\right) = \frac{1}{r} v\left(r\right) = \frac{1}{r}\left(A e^{i k r} + B e^{-i k r}\right) =\frac{1}{r} \left(C \cos\left(k r\right) + D \sin\left(k r\right)\right) \phantom{\ } \blacksquare$$

Victor Ivrii

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Re: Problem 1
« Reply #11 on: November 20, 2012, 05:18:40 AM »
http://www.math.toronto.edu/courses/apm346h1/20129/HA7.html#problem-7.1

There is an unfinished business: found solutions satisfy $\Delta u=\pm k^2u0$ as $r>0$ but not necessarily in the origin. Which solutions satisfy this equation in the origin?

Ian Kivlichan

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Re: Problem 1
« Reply #12 on: November 20, 2012, 12:20:02 PM »
http://www.math.toronto.edu/courses/apm346h1/20129/HA7.html#problem-7.1

There is an unfinished business: found solutions satisfy $\Delta u=\pm k^2u0$ as $r>0$ but not necessarily in the origin. Which solutions satisfy this equation in the origin?
I would guess that since the solutions depend on $r$ only, at $r=0$ they can only be constants $u=C$. Since they are constant, $\Delta u = 0$, so $C=0$. (Anything with $\frac{1}{r}$ dependence will have a singularity at $r=0$.)