To solve the partial differential equation [tex]\( u_x + u_y = 1 \)[/tex] using the change of variables [tex]\( s = x + y \)[/tex] and [tex]\( t = x - y \)[/tex], let's follow these steps:
1. Introduce the change of variables:
Let [tex]\( s = x + y \)[/tex] and [tex]\( t = x - y \)[/tex].
2. Express [tex]\( u \)[/tex] as a function of [tex]\( s \)[/tex] and [tex]\( t \)[/tex]:
Assume [tex]\( u \)[/tex] is now a function of [tex]\( s \)[/tex] and [tex]\( t \)[/tex], say [tex]\( u(x,y) = v(s,t) \)[/tex].
3. Calculate the partial derivatives using the chain rule:
- For [tex]\( u_x \)[/tex]:
[tex]\[
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial s} \frac{\partial s}{\partial x} + \frac{\partial v}{\partial t} \frac{\partial t}{\partial x}
\][/tex]
Since [tex]\( s = x + y \)[/tex] and [tex]\( t = x - y \)[/tex], we have:
[tex]\[
\frac{\partial s}{\partial x} = 1, \quad \frac{\partial t}{\partial x} = 1
\][/tex]
So,
[tex]\[
u_x = \frac{\partial v}{\partial s} \cdot 1 + \frac{\partial v}{\partial t} \cdot 1 = \frac{\partial v}{\partial s} + \frac{\partial v}{\partial t}
\][/tex]
- For [tex]\( u_y \)[/tex]:
[tex]\[
\frac{\partial u}{\partial y} = \frac{\partial v}{\partial s} \frac{\partial s}{\partial y} + \frac{\partial v}{\partial t} \frac{\partial t}{\partial y}
\][/tex]
Since [tex]\( s = x + y \)[/tex] and [tex]\( t = x - y \)[/tex], we have:
[tex]\[
\frac{\partial s}{\partial y} = 1, \quad \frac{\partial t}{\partial y} = -1
\][/tex]
So,
[tex]\[
u_y = \frac{\partial v}{\partial s} \cdot 1 + \frac{\partial v}{\partial t} \cdot (-1) = \frac{\partial v}{\partial s} - \frac{\partial v}{\partial t}
\][/tex]
4. Substitute the derivatives into the PDE [tex]\( u_x + u_y = 1 \)[/tex]:
[tex]\[
u_x + u_y = \left( \frac{\partial v}{\partial s} + \frac{\partial v}{\partial t} \right) + \left( \frac{\partial v}{\partial s} - \frac{\partial v}{\partial t} \right) = 1
\][/tex]
Simplifying, we get:
[tex]\[
2 \frac{\partial v}{\partial s} = 1
\][/tex]
[tex]\[
\frac{\partial v}{\partial s} = \frac{1}{2}
\][/tex]
5. Integrate with respect to [tex]\( s \)[/tex]:
[tex]\[
v(s,t) = \frac{1}{2} s + g(t)
\][/tex]
where [tex]\( g(t) \)[/tex] is an arbitrary function of [tex]\( t \)[/tex].
6. Express [tex]\( u \)[/tex] in terms of the original variables [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
Recall [tex]\( s = x + y \)[/tex] and [tex]\( t = x - y \)[/tex], so:
[tex]\[
u(x,y) = v(s,t) = \frac{1}{2} (x + y) + g(x - y)
\][/tex]
The general solution to the partial differential equation [tex]\( u_x + u_y = 1 \)[/tex] is:
[tex]\[
u(x,y) = \frac{1}{2} (x + y) + g(x - y)
\][/tex]
where [tex]\( g \)[/tex] is an arbitrary function of [tex]\( (x - y) \)[/tex].
