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PDE General Solutions

  • Thread starter erok81
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  • #1
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Homework Statement



I am not even sure if the title is correct - it's day two of the class and I am already lost beyond belief. Anyway...here is the question.

Consider the equation (1) [tex]\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} = 0[/tex]

Where u=u(x,t) is the unknown function. If f is any differentiable function of a single variable and we set u(x,t)=f(x-t) then u is a solution of (1). I'm not sure this part is even needed. :confused:

Which solution of (1) is equal to xe^(-x^2) on the x-axis?

Homework Equations



n/a

The Attempt at a Solution



I have zero idea of where to start. Well, I could probably guess a solution from my ODE knowledge, but that isn't why I am taking this course. :)

Any tips, pointers, etc?
 

Answers and Replies

  • #2
Gib Z
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On the x axis, t=0, so you are told [tex] u(x,0) = f(x) = x e^{-x^2} [/tex]. So now you know what f is. Can you find u(x,t) now?
 
  • #3
hunt_mat
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Use the method of characteristics on this equation.
 
  • #4
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Oh wow...maybe I need to drop this class and retake ODE. I still can't solve it.

Guessing a solution from ODE's I can get...

y"+2xy'+y=0. That would give me two solutions; e^(-x^2) and xe^(-x^2).

But I am pretty sure that isn't close either.
 
  • #5
hunt_mat
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Method of characteristics is for PDEs, specifically hyperbolic first order PDEs.

Search on the web
 
  • #6
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Is that the only way to solve it? Method of characteristics doesn't start until problem 9, according to my text.
 
  • #7
hunt_mat
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Is he saying that f(x-t) is a solution of your problem?
 
  • #8
HallsofIvy
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Homework Statement



I am not even sure if the title is correct - it's day two of the class and I am already lost beyond belief. Anyway...here is the question.

Consider the equation (1) [tex]\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} = 0[/tex]

Where u=u(x,t) is the unknown function. If f is any differentiable function of a single variable and we set u(x,t)=f(x-t) then u is a solution of (1). I'm not sure this part is even needed. :confused:
Yes, that's true and very important. For any differentiable function, f, the derivative of f(x- t) with respect to x, by the chain rule, is the usual derivative of f, evaluated at x-t, times the derivative of x- t with respect to x, which is 1: [itex]f_x(x- t)= f'(x- t)[/itex].

Similarly, the derivative of f(x- t) with respect to t, is the usual derivative of f, evaluated at x- t, times the derivative of x- t with respect to t, which is -1: [itex]f_t(x-t)= -f'(x- t)[/itex].
That is, [itex]u_t+ u_x= -f'(x-t)+ f'(x- t)= 0[/itex].

Because this is a first degree, in both x and t, equation, just as the general solution of a first degree ordinary differential equation depends on a single arbitrary constant, the general solution to this partial differential equation depends on a single arbitrary function.

That is, the general solution to this differential equation is [itex]u(x,y)= f(x- t)[/itex] where f can be any differentiable function of a single variable.

Which solution of (1) is equal to xe^(-x^2) on the x-axis?
On the x-axis, t= 0 so \
[tex]u(x,0)= f(x- 0)= f(x)= xe^{-x^2}[/tex]

That immediately tells you that
[tex]u(x,t)= f(x- t)= (x- t)e^{-(x- t)^2}[/tex].

Homework Equations



n/a

The Attempt at a Solution



I have zero idea of where to start. Well, I could probably guess a solution from my ODE knowledge, but that isn't why I am taking this course. :)

Any tips, pointers, etc?
 
  • #9
464
0
Ooh, that makes sense now. That you for the help. I'm not the best at showing/proofs of concepts, so it made this harder than it should have been I guess.

On a side note: I am two days into the class and this was my second problem - how bad is it that it took me a while to get? AKA how easy was that problem?
 
  • #10
Gib Z
Homework Helper
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4
Why did everyone ignore my post? What I said was pretty much spot on.

On the side note: Yes it indicates you need to concentrate more when doing problems, and get some more practice. Notice all the theory specific to Partial Differential Equations was developed for you, and to finish the question off required nothing about PDE's, just basic algebra.
 
  • #11
464
0
Why did everyone ignore my post? What I said was pretty much spot on.

On the side note: Yes it indicates you need to concentrate more when doing problems, and get some more practice. Notice all the theory specific to Partial Differential Equations was developed for you, and to finish the question off required nothing about PDE's, just basic algebra.
Now that I've done a few more problems, I realize I should have been able to solve it from what you posted and also should have been pretty simple.

I didn't ignore it, I was just still too clueless to solve from there. :redface:
 
  • #12
hunt_mat
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I think you'll find PDEs very interesting. Never underestimate the usefulness of first order PDEs, they can be very useful. I have been doing some modelling with them recently and getting some very good results.
 

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