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Let $v$ be any solution of the wave equation in double-null coordinates: $v_{x t}=0$. Show that the two equations: $$ u_x+v_x=\sqrt{2} \exp \left(\frac{u-v}{2}\right), \quad u_t-v_t=\sqrt{2} \exp \left(\frac{u+v}{2}\right), $$ are compatible iff $u$ satisfies Liouville's equation $u_{x t}=e^u$. These equations constitute a Bäcklund transformation. By considering the most general form of $v=v(x, t)$, show that: $$ u(x, t)=2 \log \left(-\frac{\sqrt{2}}{\int^x \exp [-f(\xi)] d \xi+\int^t \exp [g(\tau)] d \tau}\right)+g(t)-f(x) $$

I did the first part just fine. However, I don't quite understand how to do the last part. I know that the most general solution for $v$ is $$ v= f(x)+g(t). $$ I can plug this in and bash through the algebra to get $$ u_t -g' = u_x e^{g'/2 + f'/2} + f' e^{g'+f'}, $$ where $'$ are with respect to the natural variable. This can theoretically be solved via the method of characteristics, but it looks bad, very bad. The main issue is that there is no initial condition to tie things together, moreover, the integrals look bad and far away from the desired answer.

Question: How does one proceed for the last part? Hints/ideas are appriciated.

Maths Wizzard
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1 Answers1

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Let's start with the equations $$ u_x+v_x = \sqrt{2} \exp \left(\frac{u-v}{2}\right), \quad u_t-v_t = \sqrt{2} \exp \left(\frac{u+v}{2}\right). \tag{1} $$ Dividing the first by the second, and substituting $v$ with $f(x)+g(t)$, we obtain $$ \frac{u_x+f'(x)}{u_t-g'(t)}=e^{-f(x)-g(t)} $$ $$ \implies u_x e^{f(x)}-u_te^{-g(t)}=-f'(x)e^{f(x)}-g'(t)e^{-g(t)}. \tag{2} $$ The PDE $(2)$ can be solved using the method of characteristics. The Lagrange-Charpit equations are $$ \frac{dx}{e^{f(x)}}=-\frac{dt}{e^{-g(t)}}=-\frac{du}{f'(x)e^{f(x)}+g'(t)e^{-g(t)}}. \tag{3} $$ The solution to the first equality can be written as $$ w(x,t):=\int e^{-f(x)}dx+\int e^{g(t)}dt = C_1. \tag{4} $$ The first two terms of $(3)$ can be combined to form the equation $$ \frac{f'(x)dx-g'(t)dt}{f'(x)e^{f(x)}+g'(t)e^{-g(t)}}= -\frac{du}{f'(x)e^{f(x)}+g'(t)e^{-g(t)}}, \tag{5} $$ whose solution is $$ u(x,t)=g(t)-f(x)+C_2. \tag{6} $$ Combining $(4)$ and $(6)$, we can write the general solution to the PDE $(2)$ as $$ u(x,t)=g(t)-f(x)+F(w(x,t)), \tag{7} $$ where $F$ is, for now, an arbitrary differentiable function, and $w$ is the function in $(4)$.

To determine $F$, let's now multiply the equations $(1)$: $$ (u_x+v_x)(u_t-v_t)=2e^{u} $$ $$ \implies e^{-f(x)}F'(w)\,e^{g(t)}F'(w)=2e^{-f(x)+g(t)+F(w)} $$ $$ \implies F'(w)=\sqrt{2}\exp\left(\frac{1}{2}F(w)\right). \tag{8} $$ With the substitution $F(w)=2\log G(w)$, $(8)$ becomes $$ \frac{2G'}{G}=\sqrt{2}G \implies \frac{G'}{G^2}=\frac{1}{\sqrt{2}} \implies -\frac{1}{G}=\frac{w}{\sqrt{2}}+C. \tag{9} $$ Absorbing the constant of integration $C$ in the definition of $w$, and solving for $G$, we obtain $$ F(w)=2\log\left(-\frac{\sqrt{2}}{w}\right). \tag{10} $$ Plugging $(4)$ and $(10)$ into $(7)$, we finally arrive at $$ u(x,t)=g(t)-f(x)+2\log\left(-\frac{\sqrt{2}}{\int e^{-f(x)}dx+\int e^{g(t)}dt}\right). \tag{11} $$

Gonçalo
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