Mathematics 373 – Applied Mathematics (PDE) Solution for Problem C, PS 1

The problem was to solve the Cauchy problem

\{ \begin{matrix}u_{x} + 2u_{y} + u = e^{3x−2y} \\ u(x, 0) = 0 \end{matrix}

The solution of this inhomogeneous linear 1st order PDE has the form u = u^{p} + u^{h}, where u^{h} is the general solution of the associated homogeneous equation u_{x} + 2u_{y} + u = 0, and u^{p} is any particular solution.

We find u^{h} by changing variables:

\{ \begin{matrix}r = 2x − y \\ s = x + 2y \end{matrix}

The homogeneous equation becomes 5u_{s} + u = 0, so

u = g(r)e^{−s/5} = g(2x − y)e^{−(x+2y)/5}

where g is an arbitrary function of one variable.

To find u^{p}, we can use the idea of undetermined coefficients. e^{3x−2y} is a solution of the homogeneous equation, so we must try a function such as u^{p} = A(3x − 2y)e^{3x−2y}. Then

\begin{aligned}u_{x}^{p} + 2u_{y}^{p} + u^{p} & = e^{3x−2y}(3A + 3A(3x − 2y) − 4A − 4A(3x − 2y) + A(3x − 2y)) \\ & = −Ae^{3x−2y} = e^{3x−2y} \end{aligned}

when A = −1. This gives a general solution

u = g(2x − y)e^{−(x+2y)/5} − (3x − 2y)e^{3x−2y}.

Finally, we want u(x, 0) = 0, so

0 = g(2x)e^{−x/5} − 3xe^{3x} ⇒ g(2x) = 3xe^{16x/5}

This says

g(x) = \frac{3}{2}xe^{8x/5}

so

u = \frac{3}{2}(2x − y)e^{(16x−8y)/5}e^{(−x+2y)/5} − (3x − 2y)e^{3x−2y}

= (3x − \frac{3y}{2} ) e^{3x−2y} − (3x − 2y)e^{3x−2y}

= \frac{y}{2} e^{3x−2y}