Linear First Order PDE

  • Thread starter MisterX
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  • #1
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Homework Statement



Mod note: Pasted the OP's correction into the original problem.
Solve
[itex]xe^z\frac{\partial u}{\partial x} - 2ye^z\frac{\partial u}{\partial y} + \left(2y-x \right)\frac{\partial u}{\partial z} = 0[/itex]
given that for [itex]x > 0[/itex], [itex]u = -x^{-3}e^z[/itex] when [itex]y=-x[/itex]

Homework Equations


[itex]\mathbf{a}\left(\mathbf{x} \right) \cdot \nabla u = 0[/itex]
means that on a curve parameterized by t where
[itex]\frac{\partial \mathbf{x}}{\partial t} = \mathbf{a}\left(\mathbf{x} \right)[/itex]
[itex]\frac{\partial u}{\partial t} = \nabla u \cdot \frac{\partial \mathbf{x}}{\partial t} = \mathbf{a}\left(\mathbf{x} \right) \cdot \nabla u = 0[/itex]
In other words solutions to [itex]\frac{\partial \mathbf{x}}{\partial t} = \mathbf{a}\left(\mathbf{x} \right)[/itex], called characteristics, are the level curves of [itex]u[/itex]

The Attempt at a Solution


So the characteristic system is
[itex]\dot{x} = xe^z[/itex]

[itex]\dot{y} = -2ye^z[/itex]

[itex]\dot{z} = 2y-x[/itex]

First we can notice
[itex]e^z = \frac{\dot{x}}{x} = -\frac{1}{2} \frac{\dot{y}}{y}[/itex]
By substitution we can arrive at
[itex]\int \frac{dx}{x} = -\frac{1}{2}\int \frac{dy}{y}[/itex]
[itex]ln(x) = -\frac{1}{2}ln(y) + C[/itex]
[itex]x = C_1y^{-1/2}[/itex]
Now we may also notice that [itex]\dot{z}[/itex] looks at lot like the sum of the coefficients of [itex]e^z[/itex] in [itex]\dot{x}[/itex] and [itex]\dot{y}[/itex].
[itex]\frac{\partial}{\partial t} \left(x + y \right) = -\frac{\partial z}{\partial t}e^z = - \frac{\partial e^z}{\partial t}[/itex]
thus
[itex]e^z = x + y + C_2 = C_1y^{-1/2} + y + C_2[/itex]
[itex]z = ln\left( C_1y^{-1/2} + y + C_2\right)[/itex]
I think it means I have solved the characteristic.
[itex]y = t[/itex]
[itex]x = \frac{C_1}{\sqrt{t}}[/itex]
[itex]z = ln\left( C_1t^{-1/2} + t + C_2\right)[/itex]

However I am not sure what to do now to utilize the initial conditions and come up with the solution of the PDE. I know that the value of u at a point will be equal to the value of the initial conditions at the intersection of the initial conditions with the characteristic that goes through that point. I think we were taught about characteristics not being parallel to the initial conditions curve, but I'm not firm on exactly what the existence conditions were. Any help would be appreciated. If anyone has a suggestion for a book that covers this that might be easier or more helpful than Evans, I'd be interested.
 
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Answers and Replies

  • #2
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You went to a lot of effort to format this using LaTeX, but right off the bat, your first equation is incomprehensible.

With respect to which variables are your three partials taken?

Homework Statement


Solve
[itex]xe^z\frac{\partial u}{\partial} - 2ye^z\frac{\partial u}{\partial} + \left(2y-x \right)\frac{\partial u}{\partial} = 0[/itex]
 
  • #3
764
71
Oops, here you go.

Homework Statement


Solve
[itex]xe^z\frac{\partial u}{\partial x} - 2ye^z\frac{\partial u}{\partial y} + \left(2y-x \right)\frac{\partial u}{\partial z} = 0[/itex]
given that for [itex]x > 0[/itex], [itex]u = -x^{-3}e^z[/itex] when [itex]y=-x[/itex]
 
  • #4
haruspex
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I would set t = x, to avoid the square root.
At what value of t does a level curve intersect y = -x? What is the value of u there?
 

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