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Solving $\begin{cases} 4u_{tt}-3u_{xt}-u_{xx}=0\tag1\\u(x,0)=x^2\quad\text{and}\quad u_t(x,0)=e^x\end{cases}$ in $\mathbf R\times\mathbf R_{>0}$

First I factorized and get for the first line;

$(\partial_x-4\partial_t)(\partial_x+\partial_t)u=0$ and set $v:=(\partial_x+\partial_t)u$

then $v_x-4v_t=0$ is transport equation with solution $f\left(x+\frac{1}{4}t\right)$

hence $(1)$ is reduced to $u_x+u_t=f\left(x+\frac{1}{4}t\right)$. This is nonhomogeneous transport equation with solution

$$u(x,t)=\displaystyle\int_0^t f\left(x+\frac{1}{4}t+2(s-t)\right)ds+g(x-t)$$

considering initial conditions I get $g(x)=x^2,\ $ $\ f(x)=e^x-2x$

then wolframalpha yields, this

but if I apply $(1)$ to the result it is not true, where is the mistake ?

mathreadler
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user1161
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  • Couldn't you go from the factorisation to noting that $u$ should be of the form $f(4x+t)+g(x-t)$, and then applying the conditions? – πr8 May 01 '16 at 11:28
  • @πr8 Yes that is another method, I read it somewhere but we didn't treat the classificiation of the pdes (as hyperbolic, elliptic, parabolic) so I think I have to use this – user1161 May 01 '16 at 11:31

1 Answers1

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$(\partial_x-4\partial_t)(\partial_x+\partial_t)u=0$ and set $v:=(\partial_x+\partial_t)u$

then $v_x-4v_t=0$ is transport equation with solution $f\left(x+\frac{1}{4}t\right)$

Hence the problem is reduced to $u_x+u_t=f\left(x+\frac{1}{4}t\right)$.

This is the nonhomogeneous transport equation with solution

$u(x,t)=\int_0^t f(x+s-t+\frac 14 s)ds+g(x-t)\tag{$\star$}$

Initial condition $u(x,0)$ implies $g(x)=x^2$

second initial condition $u_t(x,0)$ yields

$f(x)=e^x+2x$

Inserting in $(\star)$ gives,

This

user1161
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