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.


$$3 \frac{\partial{u}}{\partial{t}} + t^2\frac{\partial{u}}{\partial{x}} = 0$$ 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 : $$u(x,t)=f\left(x-\frac{1}{9}t^3\right)$$

  • $\begingroup$ 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

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