finding a solution of this PDE


by climbon
Tags: solution
climbon
climbon is offline
#1
Jul28-11, 06:33 AM
P: 16
Hi,

i'm having trouble finding a solution to this PDE,

[tex] \frac{d U(x,y,t)}{dt} = A(x) \frac{\partial U(x,y,t)}{\partial y} + B(y) \frac{\partial U(x,y,t)}{\partial x}[/tex]

with only knowledge of the initial condition U(x,y,0)=F(x,y).

I've tried to solve this using characteristics but the only examples i can find in books is for the case when the left hand side is zero. Tried following the method from some books but can only solve it for when the L.H.S is zero. I'm not sure where to go next

Any help would be fantastic.

Thanks.
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hunt_mat
hunt_mat is offline
#2
Jul28-11, 12:52 PM
HW Helper
P: 1,584
For the LHS do you mean:
[tex]
\frac{\partial U}{\partial t}
[/tex]
matphysik
matphysik is offline
#3
Jul28-11, 08:13 PM
P: 85
Quote Quote by climbon View Post
Hi,

i'm having trouble finding a solution to this PDE,

[tex] \frac{d U(x,y,t)}{dt} = A(x) \frac{\partial U(x,y,t)}{\partial y} + B(y) \frac{\partial U(x,y,t)}{\partial x}[/tex]

with only knowledge of the initial condition U(x,y,0)=F(x,y).

I've tried to solve this using characteristics but the only examples i can find in books is for the case when the left hand side is zero. Tried following the method from some books but can only solve it for when the L.H.S is zero. I'm not sure where to go next

Any help would be fantastic.

Thanks.
The LHS should read ∂U(x,y,t)/∂t not dU(x,y,t)/dt.

Then, for (parameter) s∈I⊂ℝ:

d/ds[U(x(s),y(s),t(s))]= ∂U/∂xdx/ds + ∂U/∂ydy/ds + ∂U/∂tdt/ds≡ B(y)∂U/∂x + A(x)∂U/∂y - ∂U/∂t= 0.

You seek, U(x(s),y(s),t(s))= constant.

ADDENDUM:
Hint: dx/B = dy/A = dt/-1.


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