To reduce the partial differential equation
[tex]\[
a u_x + b u_y + c u = 0
\][/tex]
to a first-order ordinary differential equation by introducing the change of variables [tex]\( s = ax + by \)[/tex] and [tex]\( t = bx - ay \)[/tex], we proceed as follows:
### Step 1: Express [tex]\( u \)[/tex] in terms of the new variables [tex]\( s \)[/tex] and [tex]\( t \)[/tex]
Given the function [tex]\( u = u(s, t) \)[/tex], where [tex]\( s = ax + by \)[/tex] and [tex]\( t = bx - ay \)[/tex].
### Step 2: Compute the partial derivatives of [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
Using the chain rule for partial derivatives, we have:
[tex]\[
u_x = \frac{\partial u}{\partial s} \frac{\partial s}{\partial x} + \frac{\partial u}{\partial t} \frac{\partial t}{\partial x}
\][/tex]
[tex]\[
u_y = \frac{\partial u}{\partial s} \frac{\partial s}{\partial y} + \frac{\partial u}{\partial t} \frac{\partial t}{\partial y}
\][/tex]
First, compute the needed derivatives:
[tex]\[
\frac{\partial s}{\partial x} = a, \quad \frac{\partial s}{\partial y} = b
\][/tex]
[tex]\[
\frac{\partial t}{\partial x} = b, \quad \frac{\partial t}{\partial y} = -a
\][/tex]
Thus:
[tex]\[
u_x = \frac{\partial u}{\partial s} a + \frac{\partial u}{\partial t} b
\][/tex]
[tex]\[
u_y = \frac{\partial u}{\partial s} b - \frac{\partial u}{\partial t} a
\][/tex]
### Step 3: Substitute these expressions into the original PDE
The original PDE in terms of [tex]\( u_x \)[/tex] and [tex]\( u_y \)[/tex] is:
[tex]\[
a u_x + b u_y + c u = 0
\][/tex]
Substitute the expressions for [tex]\( u_x \)[/tex] and [tex]\( u_y \)[/tex]:
[tex]\[
a (a u_s + b u_t) + b (b u_s - a u_t) + c u = 0
\][/tex]
### Step 4: Simplify the equation
[tex]\[
a^2 u_s + ab u_t + b^2 u_s - ab u_t + c u = 0
\][/tex]
Notice that the [tex]\( u_t \)[/tex] terms cancel out:
[tex]\[
(a^2 + b^2) u_s + c u = 0
\][/tex]
### Step 5: Collect terms involving [tex]\( u \)[/tex] and [tex]\( u_s \)[/tex]
The equation simplifies to:
[tex]\[
(a^2 + b^2) \frac{\partial u}{\partial s} + c u = 0
\][/tex]
### Step 6: Solve this as a first-order ODE
This is now a first-order ordinary differential equation in terms of [tex]\( s \)[/tex]:
[tex]\[
(a^2 + b^2) \frac{du}{ds} + c u = 0
\][/tex]
If we divide the entire equation by [tex]\( a^2 + b^2 \)[/tex] (assuming [tex]\( a^2 + b^2 \ne 0 \)[/tex]):
[tex]\[
\frac{du}{ds} + \frac{c}{a^2 + b^2} u = 0
\][/tex]
This is a linear first-order ordinary differential equation of the form:
[tex]\[
\frac{du}{ds} + \lambda u = 0
\][/tex]
where [tex]\( \lambda = \frac{c}{a^2 + b^2} \)[/tex].
### Step 7: Solution of the first-order ODE
The solution to this first-order linear ODE is:
[tex]\[
u(s) = u_0 \exp \left(-\lambda s \right) = u_0 \exp \left(-\frac{c}{a^2 + b^2} s \right)
\][/tex]
where [tex]\( u_0 \)[/tex] is an integration constant that may depend on the second variable [tex]\( t \)[/tex]:
[tex]\[
u(s, t) = u_0(t) \exp \left(-\frac{c}{a^2 + b^2} s \right)
\][/tex]
Thus, the partial differential equation [tex]\( a u_x + b u_y + c u = 0 \)[/tex] has been reduced to a first-order ordinary differential equation as outlined above.
