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PDE concept Help |
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| Sep9-10, 09:29 PM | #1 |
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PDE concept Help
1. The problem statement, all variables and given/known data
Uxx + 3*Uyy - 2*Ux + 24*Uy +5*U = 0 Reduce this to the form Vxx + Vyy + C*V = 0 U = V*e^(alpha*x + Beta*y) y' = gamma*y Ok, the problem I am having is I don't know what to do with the gamma, however I am off by a factor of 3 in my answer for Vyy, so I know gamma must be 1/sqrt(3). I just don't understand how the gamma works here. 2. Relevant equations I am not quite sure what to put in this space, first time posting, and have everything up to I believe 3. The attempt at a solution Ok, I have simply taken partial derivatives of the equation U = V*e^(alpha*x + Beta*y) This has churned out the following equation when plugging the partial derivatives back into the PDE above: Vxx*e^(alpha*x + Beta*y) + 3*Vyy*e^(alpha*x + Beta*y) + (alpha -1)*Vx*e^(alpha*x + Beta*y) + (6*Beta + 24)*Vy*e^(alpha*x + Beta*y) + ((alpha)^2 + (Beta)^2 + 24*Beta - 2*alpha + 5)*V*e^(alpha*x + Beta*y) = 0 as seen, there is a factor of 3 I can't account for because it is easily found that alpha = 1 and Beta = -4. I know the 3 is killed by gamma = 1/sqrt(3), however IDK how it works. The equation is then reduced to (I am leaving the e^(alpha*x + Beta*y) in although it could be divided out easily): Vxx*e^(alpha*x + Beta*y) + 3*Vyy*e^(alpha*x + Beta*y) + ((alpha)^2 + (Beta)^2 + 24*Beta - 2*alpha + 5)*V*e^(alpha*x + Beta*y) = 0 I would really appreciate an explanation for why and how this gamma works. Thanks! |
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