5

It is asked to solve the PDE

$$u_{xx} - 3u_{xt} - 4u_{tt} = 0$$ using a factorization, that consists in

$$\left( \frac{\partial}{\partial x} - 4 \frac{\partial}{\partial t} \right) \left( \frac{\partial}{\partial x} + \frac{\partial}{\partial t} \right) u = 0$$

My attempt:

I called

$$\left( \frac{\partial}{\partial x} + \frac{\partial}{\partial t} \right) = v$$ and solved $$ v_x - 4 v_t = 0$$ making the change of variables: $x' = x-4t, \; t' = -4x - t$. This change of variables leave us with the following "ODE": $$17 u_{x'} = 0 \Rightarrow u(x,t) = f(t') = f(-4x-t) = g(4x+t)$$ where g and f are arbitrary functions of one variable

For the second part, we have the PDE: $$u_x+u_t = g(4x+t)$$

I made the following change of variables: $x' = x+t, t' = x-t$, and this leave us with the following "ODE": $$2u_{x'} = g(4x+t)$$

The answer is $$ u(x,t) = c_1(4x+t)+c_2(x-t)$$

I understand that the last "ODE" give us the solution

$$u(x,t) = \int g(4x+t) dx' + c_2(x-t)$$ but why is the integral also a function of 4x+t? Where I commited a mistake?

Thanks in advance!

@Edit: Please, don't give me a solution using another technique that is not a factorization. I think my mistake, if there is one, consists on the variables that I am choosing.

Giiovanna
  • 3,197
  • You are already done as $c_1(4x+t)=\int g(4x+t)dx'=\int g(\frac52x'+\frac32t')dx'$. If $G$ is an anti-derivative of $g$, then $c_1(u)=\frac25G(u)$. – Lutz Lehmann May 31 '18 at 07:01

1 Answers1

2

It will be helpful to view the "factorization" technique as separation of variables. Assume $\,x = \eta+\xi ,\,$ and $\,t = \eta-4\xi .\,$ Then

\begin{cases} x = \eta + \xi\\ t = \eta-4\xi \end{cases} \implies \begin{cases} \dfrac{\partial u}{\partial \xi} = \dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial \xi} + \dfrac{\partial u}{\partial t}\dfrac{\partial t}{\partial \xi} = \dfrac{\partial u}{\partial x} - 4 \dfrac{\partial u}{\partial t} \\ \dfrac{\partial u}{\partial \eta} = \dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial \eta} + \dfrac{\partial u}{\partial t}\dfrac{\partial t}{\partial \eta} = \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial t} \end{cases}

But then

\begin{multline} \left( \frac{\partial}{\partial x} - 4 \frac{\partial}{\partial t} \right) \left( \frac{\partial}{\partial x} + \frac{\partial}{\partial t} \right) u = \left( \frac{\partial}{\partial \xi} \right) \left( \frac{\partial}{\partial \eta} \right) u = u_{\xi\eta} = 0 \end{multline}

We can write

\begin{alignat}{1} u_{\xi\eta} = 0 & \implies u_{\xi} & = \int u_{\xi\eta} \,d\eta = c_1\left(\xi\right), \quad u_{\eta} = \int u_{\xi\eta} \,d\xi = c_2\left(\eta\right),& \\ &\implies u & = \iint u_{\xi\eta} \,d\eta \,d\xi = \int c_1\left(\xi\right) \, d\xi = C_1\left(\xi\right) + C_2\left(\eta\right)& \\ &\implies u_\eta & = \dfrac{d C_2}{d\eta} = \int u_{\xi\eta} \,d\eta = c_2\left(\eta\right) &\\ &\implies u & = C_1\left(\xi\right) + C_2\left(\eta\right)& \end{alignat}

In that case it is clear that

$$ u = C_1\left(\xi\right) + C_2\left(\eta\right) = C_1\big(x-t \big) + C_2\big(4x+t \big) $$

Since $\,c_1,\,$ $\,c_2,\,$ and $\,c_3\,$ are arbitrary functions, we can rewrite the last equation as

$$ \bbox[5pt, border:2.5pt solid #FF0000]{ u = C_1 \left(x-t \right) + C_2 \left(4x+t \right) } $$

Vlad
  • 6,710