# PDE Method of characteristics with initial condition

I wanted to solve the following PDE with initial condition $$u_t+tu_x=0,$$ $$u(x,1)=f(x),$$ where $f(x)$ is a given function, using the method of characteristics.

I explain what I have done. First of all the characteristic system is $$x'(\tau)=t$$ $$t'(\tau)=1$$ $$u'(\tau)=u$$ with initial conditions $$x(s,1)=s$$ $$t(s,1)=1$$ $$u(s,1)=f(s)$$

It follows that $$x(s,\tau)=\tau^2+s-1,$$ $$t(s,\tau)=\tau,$$ $$u(s,\tau)=e^{\tau-1}f(s).$$ Therefore, $s=x-t^2+1$ and $\tau=t$. Hence $$u(x,t)=u(x-t^2+1,t)=e^{t-1}f(x-t^2+1).$$

But this is wrong because it does not satisfy the system. What would be right is $$u(x,t)=e^{t-1}f(x-t^2/2+1/2).$$ But I do not know where this $1/2$ can come from...

Thanks in advance!

## 1 Answer

The integral of $\tau$ is $\tau^2/2$, not $\tau$. It should be $$x=\frac{\tau^2}{2}+s-\frac12.$$

• Thank you very much! This means that in the characteristic system must be $x'(\tau)=\tau$, not $x'(\tau)=t$, true? In spite of we know $t=\tau$ after solving $t'(\tau)=1$ with initial condition $t(s,1)=1$. – S. Proa Feb 1 '16 at 15:54
• Yes, that's it. – Julián Aguirre Feb 1 '16 at 17:02