How to solve the given PDE?

Question

I want to solve the given PDE: $$p_t=-(wp_y+zp_x)+B(p_{yy}+p_{xx})$$ where $p(x,y,t)$ and $B,w,z$ are constants I know how to solve the equation given by $$p_t=B(p_{yy}+p_{xx})$$ as the steps are analogus to heat equation solution. But cannot figure out the solution and the necessary steps for solving $$p_t=-(wp_y+zp_x)+B(p_{yy}+p_{xx})$$ for certain boundary conditions, preferably using Fourier series? Can somebody provide me a brief outline for the general solution?

Answer 1

Hint.

After Laplace transforming, consider the PDE

$$ s p(x,s)-p(x,0) = \partial_x(ap(x,s)+b\partial_x p(x,s)) $$

now assuming $p(x,0) = 0$ for simplicity, the resolution follows much the way an ODE

$$ p(x,s) = c_1(s) e^{\frac{1}{2} x \left(-\frac{\sqrt{a^2+4 b s}}{b}-\frac{a}{b}\right)}+c_2(s) e^{\frac{1}{2} x \left(\frac{\sqrt{a^2+4 b s}}{b}-\frac{a}{b}\right)} $$

now knowing the boundary conditions for $x$ we can determine $c_1(s), c_2(s)$ and after that we obtain the Laplace inversion: directly or with the help of residue theory.

NOTE

Considering $p(x,y,t) = X(x)Y(y)T(t)$ with the PDE

$$ p_t = b(p_{xx}+p_{yy}) $$

we have

$$ \frac{\dot T}{T} = b\left(\frac{X''}{X}+\frac{Y''}{Y}\right) = \lambda $$

and also

$$ \frac{X''}{X}+\frac{Y''}{Y} = \frac{\lambda}{b}\Rightarrow \frac{X''}{X} = \frac{\lambda}{b}-\frac{Y''}{Y} = \mu $$

so we have

$$ \dot T = \lambda T\\ X'' = \mu X\\ Y'' = \left(\frac{\lambda}{b}-\mu\right)Y $$