- #1

Emspak

- 243

- 1

## Homework Statement

OK, a PDE: $$a\frac{\partial u}{\partial t} + b \frac{\partial u}{\partial x} = u$$

we want the general solution.

**2. The attempt at a solution**

So, I'll set up a couple of equations thus:

r = m

_{11}x + m

_{21}t

s = m

_{12}x + m

_{22}t

(We have a nice matrix of m here if we want).

partially differentiating both:[itex] \frac{\partial r}{\partial x}= m_{11},\frac{\partial r}{\partial t}= m_{12}, \frac{\partial s}{\partial x}= m_{21}, \frac{\partial s}{\partial t}= m_{22} [/itex]

and plugging them back into the original PDE we get: $$ \frac{\partial u}{\partial t}= \frac{\partial u}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial u}{\partial s}\frac{\partial s}{\partial t}= m_{12} \frac{\partial u}{\partial r}+ m_{22}\frac{\partial u}{\partial s} $$ and $$ \frac{\partial u}{\partial x}= \frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}= m_{21} \frac{\partial u}{\partial s}+ m_{11}\frac{\partial u}{\partial r}=u$$

from there I substitute in to the original PDE and got: $$a\left(m_{12} \frac{\partial u}{\partial r}+ m_{22} \frac{\partial u}{\partial s} \right) + b\left(m_{11} \frac{\partial u}{\partial r}+ m_{21} \frac{\partial u}{\partial s} \right) \Rightarrow (am_{12}+bm_{11}) \frac{\partial u}{\partial r}+ (am_{22}+bm_{21}) \frac{\partial u}{\partial s} =u$$

So I want to get rid of one of the partials. The simplest way to do that is to set one of the coefficients equal to zero. So:

am

_{12}+bm

_{11}=0. That makes m

_{12}=-b and m

_{11}=a. So my r from above is r=ax-bt.

That leaves me with

[itex]\left(am_{22}+bm_{21}\right)\frac{\partial u}{\partial s}=u[/itex]

and I can set [itex]\left(am_{22}+bm_{21}\right)= 1[/itex] leaving [itex]m_{22} = \frac{1}{a}[/itex] and [itex]m_{21}=0[/itex]. From the equation for [itex]s[/itex]above that leaves us with [itex]s=\frac{1}{a}t[/itex] and we go back to the partials.

Since we set the coefficient of [itex]\frac{\partial u}{\partial s}[/itex] equal to 1 we are left with [itex]\frac{\partial u}{\partial s}=u[/itex] and that is a simple ODE that reduces to [itex]u=Ce^{-s}[/itex] and

Since $$u=f(r)e^{-s} = f(ax-bt)e^{-s}= f(ax-bt)e^{-\frac{t}{a}}$$ as our general solution.

So how did I do? We're still in the beginning of the class, and this was one of the problems that is supposed to explore "easy" methods; I know that the method of characteristics would be better.