1
$\begingroup$

I'm trying to turn the following nonlinear PDE into a nonlinear ODE using dimensional reduction (since u is a velocity) and a similarity variable.

$\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x}= 0$

With conditions:

$u(x,0) = 0$ for $x < 0$ and $u(x,0) = u_{0}$ for $x > 0$.

After dimensional reduction, I find that:

$u = u_{0}F(y)$, where $y = \frac{x}{u_{0}t}$ is my similarity variable and $F$ is yet to be determined.

However, when I plug back into the PDE to try to solve for $F$ rather than getting a nonlinear ODE, most everything just cancels out and I get that:

$F(y) = y$

which doesn't seem to satisfy the conditions. I have no idea what else to do though. Any suggestions/hints? This isn't even for a differential equations class, this is just one part of a larger problem and my Diff Eq is pretty rusty so I'm even more confused.

1 Answers 1

1

The difficulty comes from the condition $u(x,0)=u_0$ for $x>0$.

So, at the starting time $t=0$ we have $\frac{\partial u}{\partial x}=0$ since $u(x)=u_0=$constant.

Putting it into $\quad \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x}= 0\quad $ leads to $\quad\frac{\partial u}{\partial t}=0$

Thus, $u$ remains constant $\quad u(x,t)=u_0$

Alternatively, this result can be analytically obtained, thanks to the method of characteristics which leads to the general solution of the PDE on the form of an implicit equation : $$u=f(x-tu)$$ any differentiable function $f$.

The condition $u(x,0)=u_0$ implies $u_0=f(x)$ which defines the function $f$ as a constant function (not function of the variable). So, the solution according to the specified condition is : $$u(x,t)=u_0$$

Since you didn't post all the steps of your calculus, it isn't possible to see if there is or not a mistake in what you did.