du/dt-t(du/dx)=-u^2
d is meant to be partial derivatives.
I am going to try. I cannot gaurentee my solution is correct.
$\displaystyle \frac{\partial u}{\partial t} - t\cdot \frac{\partial u}{\partial x} = - u^2$
Say we want to solve this subject to some curve $\displaystyle x=x(s) \mbox{ and }t=t(s)$ with $\displaystyle x(0)=x_0 \mbox{ and }t(0)=0$.
Then $\displaystyle u(x,t) = u(x(s),t(s))$
Use the chain rule,
$\displaystyle \frac{du}{ds} = \frac{\partial u}{\partial x} \cdot \frac{dx}{ds} + \frac{\partial u}{\partial t} \cdot \frac{dt}{ds}$
By the method of charachteristics we set,
1)$\displaystyle \frac{dt}{ds} = 1 \mbox{ with }t(0)=0$
2)$\displaystyle \frac{dx}{ds} = -t \mbox{ with }x(0) = x_0$
3)$\displaystyle \frac{du}{ds} = -u^2$
The solution to the IVP problems are:
1)$\displaystyle t = s$ (**)
2)$\displaystyle x = x_0 - ts$ (*)
The solution to #3 is simple:
$\displaystyle \frac{du}{dx} = -u^2 \Rightarrow -\frac{1}{u^2}u' = 1$
This is a Seperable equation,
$\displaystyle \frac{1}{u} = s + F(x_0) \Rightarrow u =\frac{1}{s+F(x_0)}$
Where $\displaystyle F(x_0)$ is an arbitrary starting point on the solution curve.
By (*) we see that $\displaystyle x_0 = x+ts$.
Thus by (**),
$\displaystyle u(x,t) = \frac{1}{t+F(x+ts)}$
Where $\displaystyle F$ is any arbitrary differencial function (and not equal to $\displaystyle -s$).
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Which seems to work !!! Because if $\displaystyle F = 0$ everywhere then $\displaystyle u(x,t) = \frac{1}{t+0} = \frac{1}{t}$. If you substitute that into the PDE you get:
$\displaystyle \frac{\partial u}{\partial t} = - \frac{1}{t^2} = - u^2$
Which works.