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PDE cauchy question

by finmaths123
Tags: cauchy
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finmaths123
#1
Aug15-11, 06:57 AM
P: 1
Hi guys tryin to study for a pde exam and cannot solve this question


Find a general solution of the equation

exp(-x)dz/dx+{/y(squared)}dz/dy=exp(x)yz

(ii) Solve the Cauchy problem, i.e. find the integral surface of this equation
passing through the curve .
y = ex/3 , z = e .
Calculate the partial derivatives of the solution z(x, y) of the Cauchy problem you found and show that it satisfies the partial differential equation.

Help would be greatly appreachated
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snipez90
#2
Aug15-11, 09:56 PM
P: 1,104
There is no excuse for not showing some work to a problem like this. The first step is understanding what type of PDE you're actually dealing with. You can also try thinking of general methods such as "method of characteristics" or "fourier transform" (one of these may be helpful here).
jackmell
#3
Aug16-11, 07:33 AM
P: 1,666
Quote Quote by snipez90 View Post
There is no excuse for not showing some work to a problem like this. The first step is understanding what type of PDE you're actually dealing with. You can also try thinking of general methods such as "method of characteristics" or "fourier transform" (one of these may be helpful here).
You can't cut him a break snipez? It's just his first post and my understanding is that only in the homework forums are you explicitly required to show work. I'm pretty forgiving and would at least have led him on by just starting the conversation such as, "what happens when you set up the characteristic equation:"

[tex]\frac{dy}{dx}=y^2 e^x[/tex]

Personally, I'd like to see the problem worked through. I tried finding the particular solution but could not do it. But it's not my thread so I'm explicitly not asking anyone to help me. But you guys could go easy on this new guy and help him a little. :)

hunt_mat
#4
Aug16-11, 08:12 AM
HW Helper
P: 1,583
PDE cauchy question

You can use the method of characteristics to get a set of ODEs
[tex]
\dot{x}=e^{-x}, \quad\dot{y}=\frac{1}{y^{2}}, \quad\dot{z}=e^{x}y
[/tex]
You have to think about some initial conditions for this, can you suggest something?
jackmell
#5
Aug17-11, 07:06 AM
P: 1,666
Quote Quote by hunt_mat View Post
You can use the method of characteristics to get a set of ODEs
[tex]
\dot{x}=e^{-x}, \quad\dot{y}=\frac{1}{y^{2}}, \quad\dot{z}=e^{x}y
[/tex]
You have to think about some initial conditions for this, can you suggest something?
Thanks, didn't think to solve it parametrically although the text I have used two parameters. Also, I think there is some ambiguity in how the PDE was originally written. For the results here, I used:

[tex]e^{-x}u_x+\frac{1}{y^2}u_y-e^{x}yu=0;\quad u(x,ex/3)=e[/tex]

and obtained:

[tex]x(s,t)=ln(t+e^s)[/tex]

[tex]y(s,t)=\sqrt[3]{t+1/3\left(\frac{es}{3}\right)^3}[/tex]

[tex]u(s,t)=\frac{e}{\text{exp}\left\{-\int_{0}^t e^{x(s,v)}y(s,v)dv\right\}}[/tex]

Also, using the two-parameter approach, the last DE of the set is:

[tex]\frac{du}{dt}=e^x y u[/tex]

I can compare those to a NDSolve of the equations in Mathematica and they do seem to agree:

Clear[x, y, s, t, u]

mysol = NDSolve[{D[x[s, t], t] == Exp[-x[s, t]], 
    D[y[s, t], t] == 1/y[s, t]^2, 
    D[u[s, t], t] == Exp[x[s, t]]*y[s, t]*u[s, t], 
    x[s, 0.001] == s, y[s, 0.001] == E*(s/3), 
    u[s, 0.001] == E}, {x, y, u}, {t, 0.001, 1}, 
   {s, 0.01, 1}]
But I see no way to directly back-substitute the partials back into the PDE and check it directly.
snipez90
#6
Aug17-11, 08:26 PM
P: 1,104
My reply probably came off harsher than I intended it. But my understanding is that if there is a homework type question, it shouldn't be here. From the context of the OP, I gathered this was a homework problem (even though it only has to be a homework "type" problem).
hunt_mat
#7
Aug18-11, 04:52 AM
HW Helper
P: 1,583
Although the solution presented by Jackmell is technically correct, it is not particularly nice. As Snipez90 pointed out this looks to be rather a book like question and book like questions tend to have nice answers, so I think that we should get the poster to clarify the coefficient of [itex]\partial_{y}u[/itex] because the solution is much nicer for the coefficient to be [itex]y^{2}[/itex] than [itex]y^{-2}[/itex]. Either that or we are missing a trick somewhere.


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