PDE trouble - I'm a little rusty

• Aug 8th 2010, 06:23 AM
funnyinga
PDE trouble - I'm a little rusty
Hello,
Its been a long time since I've done PDE's or even ODE's for that matter, and I'm quite rusty. Could anyone please help me with the following?

Solve the following PDE and explain why inverting the equations for \$\displaystyle x(s, t)\$ and \$\displaystyle y(s,t)\$ may cause problems:

\$\displaystyle u_x + u_y = u\$ subject to \$\displaystyle u(x, ax) = F(x) \$

Find the Jacobian of the coordinate transform to determine when the transformation is invertible.
• Aug 8th 2010, 06:51 AM
Jester
If we let

\$\displaystyle u_s = u_x x_s + u_y y_s\$

then choosing

\$\displaystyle x_s = 1, y_s = 1\;\;(1)\$ gives \$\displaystyle u_s = u_x + u_y = u \;\;(2)\$ (your PDE)

The boundary conditions. If we let \$\displaystyle y = ax\$ be \$\displaystyle s = 0\$ and \$\displaystyle x = r\$ then we have

On \$\displaystyle s = 0, x = r, y = ar \$ and \$\displaystyle u = F(r)\;\;(3)\$

Integrating (1) and (2) gives

\$\displaystyle x = s + A(r),\;\;\;y = s + B(r),\;\;\;u = C(r)e^s\$

where \$\displaystyle A, B\$ and \$\displaystyle C\$ are arbitrary. Imposing the BCs (3) gives

\$\displaystyle x = s + r,\;\;\; y = s + ar,\;\;\;u=F(r)e^s,\;\;(4)\$

noting Jacobian of the transformation for \$\displaystyle x \$ and \$\displaystyle y\$ vanishes if \$\displaystyle a = 1.\$. Now solving the first two of (4) for r and s and substituting into the third of (4) gives your solution.