Question

Show that each of the following equations has a solution of the form u(x,y) = e^(ax+by). Find the constants a,b for each example:

a) u_x + 3u_y + u = 0.

b) u_xx + u_yy = 5e^(x-2y).

Answer 1

Show that each of the following equations has a solution of the form

$$u(x, y) = e^{a x + b y}.$$

Find the constants $a, b$ for each example:

a) $u_x + 3u_y + u = 0$.

b) $u_{xx} + u_{yy} = 5e^{x - 2y}$.

Solution:

a)

We have

$$u_x(x, y) = \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(e^{a x + b y}) = a e^{a x + b y};$$ $$u_y(x, y) = \frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(e^{a x + b y}) = b e^{a x + b y}.$$

Then

$$u_x + 3u_y + u = 0,$$ $$a e^{a x + b y} + 3b e^{a x + b y} + e^{a x + b y} = 0,$$ $$(a + 3b + 1) e^{a x + b y} = 0.$$

So the function $u(x, y) = e^{a x + b y}$ will be the solution of the equation $u_x + 3u_y + u = 0$ if

$$a + 3b + 1 = 0,$$ $$a = -3b - 1, \quad b \in \mathbb{R}.$$

b)

We have

$$u_{xx}(x, y) = \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right) = \frac{\partial}{\partial x}(a e^{a x + b y}) = a^2 e^{a x + b y};$$ $$u_{yy}(x, y) = \frac{\partial}{\partial y}\left(\frac{\partial u}{\partial y}\right) = \frac{\partial}{\partial y}(b e^{a x + b y}) = b^2 e^{a x + b y}.$$

Then

$$u_{xx} + u_{yy} = 5e^{x - 2y},$$ $$a^2 e^{a x + b y} + b^2 e^{a x + b y} = 5e^{x - 2y},$$ $$(a^2 + b^2) e^{a x + b y} = 5e^{x - 2y}.$$

So the function $u(x, y) = e^{a x + b y}$ will be the solution of the equation $u_{xx} + u_{yy} = 5e^{x - 2y}$ if

$$\left\{ \begin{array}{c} a^2 + b^2 = 5, \\ a x + b y = x - 2y. \end{array} \right.$$

So we have $a = 1$, $b = -2$.

Answer:

a) $a = -3b - 1$, $b \in \mathbb{R}$.

b) $a = 1$, $b = -2$.