Answer to Question #149760 in Differential Equations for Ashweta Padhan

Question

uxx - (sech^4x)uyy = 0
Reduce into canonical form and solve it

Answer 1

$u_{xx}-sech^4xu_{yy}=0$

$A=1, B=0, C=-sech^4x$

$\Delta=B^2-4AC=4sech^4x>0$

This is hyperbolic PDE.

The characteristic polynomial of the PDE:

$A(\frac{dy}{dx})^2-B(\frac{dy}{dx})+C=0$

$\frac{dy}{dx}=\frac{B\pm\sqrt{B^2-4AC}}{2A}=\pm sech^2x$

$y=tanhx+k_1$

$y=-tanhx+k_2$

$\xi=y-tanhx$

$\eta=y+tanhx$

We have:

$a=0, c=0$

Canonical form:

$\omega_{\xi\eta}=0$

$\omega(\xi,\eta)=f(\xi)+g(\eta)$

The general solution:

$u=f(y-tanhx)+g(y+tanhx)$