1
$\begingroup$

Consider the PDE $$3 \frac{\partial{u}}{\partial{t}} + t^2\frac{\partial{u}}{\partial{x}} = 0: t \lt 0 $$

with the initial condition $$u(x,0) = f(x): 0 \lt x \lt 1$$

Determine the characteristics and sketch them.

I understand that the PDE is of the semi-linear homogenous form but i'm struggling with the method.

So far I have $\frac{\partial{t}}{\partial{\sigma}} = 3$, $\frac{\partial{x}}{\partial{\sigma}} = t^2$ and $\frac{\partial{z}}{\partial{\sigma}} = 0$.

$\Rightarrow t = 3\sigma + t_0, x = t^2\sigma+ x_0, z = z_0$.

But I don't understand how to proceed and incorporate the initial conditions too.

$\endgroup$
1
$\begingroup$

$$3 \frac{\partial{u}}{\partial{t}} + t^2\frac{\partial{u}}{\partial{x}} = 0$$ The characteristic relationships are : $$\frac{dt}{3}=\frac{dx}{t^2}=\frac{du}{0}$$ A first characteristic results from $\frac{dt}{3}=\frac{dx}{t^2} \qquad\to\qquad dx-\frac{1}{3}t^2 dt=0$ $$x-\frac{1}{9}t^3=c_1$$ A second characteristic results from necessarily $du=0$ : $$u=c_2$$ Thus, the general solution expressed on implicite form is : $$\Phi\left(x-\frac{1}{9}t^3\:,\:u\right)=0$$ where $\Phi$ is any differentiable function of two variables.

This implicite equation can be solved for $u$ : $$u=F\left(x-\frac{1}{9}t^3\right)$$ where $F$ is any differentiable function of one variable.

The boundary condition : $$u(x,0)=f(x)$$ where $f(x)$ is a given function, leads to : $$f(x)=F\left(x-\frac{1}{9}0^3\right)=F(x)$$ which determines $F(x)=f(x)$

So, the final result is : $$u(x,t)=f\left(x-\frac{1}{9}t^3\right)$$

$\endgroup$
  • $\begingroup$ Thanks for the in-depth and clear solution. I'm just curious as to how the implicite equation would be solved for u? $\endgroup$ – Goshawk Apr 15 '16 at 13:29
  • $\begingroup$ In such an equation $\Phi(X,Y)=0$ since $\Phi$ is any function, one variable is any function of the other. For example $Y=F(X)$ any function $F$. $\endgroup$ – JJacquelin Apr 15 '16 at 13:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.