PDE Method of characteristics with initial condition

Question

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!

score 2 · Accepted Answer · answered Feb 01 '16 at 14:36

2

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

answered Feb 01 '16 at 14:36

Julián Aguirre

76,354

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 01 '16 at 15:54
Yes, that's it. – Julián Aguirre Feb 01 '16 at 17:02

PDE Method of characteristics with initial condition

1 Answers1