How solve the following PDE?

Question

I want to solve the following PDE:$$ \begin{cases} u_{tt}-c^2u_{xx}=0, \quad x\in\mathbb{R},\ t\geq x\\ u(x,x)=φ(x), \quad x\in\mathbb{R}\\ u_t(x,x)=0, \quad x\in\mathbb{R} \end{cases} $$ where $φ:\mathbb{R} \to \mathbb{R}$, $φ\in C^1(\mathbb{R})$.

Thanks for your help.

As a first try, I would separate the variables writing $u(t,x) = a(t)b(x)$. If you replace this $u$ in the equation, after simple steps you may write a left hand side depending only on $t$, and the right hand side depending only on $x$. This means both sides must be constant. So you find two simpler equations $a''(t) = \alpha a(t)$ and $b''(x) = \alpha c^{-2} b(x)$, and the solutions of these are classified (they are linear ODEs). Watch out: I divided by $c$, so treat the case $c = 0$ separately. — Gibbs, Apr 13 '18 at 22:26
Yes. You may assume the solution to be of the form $u(t,x) = a(t)b(x)$ and see what happens. — Gibbs, Apr 13 '18 at 22:36
$u(t,x) = a(t)b(x)$, then $u_t(t,x) = a_t(t)b(x)$, $u_t(x,x) = a_t(x)b(x)=0$, $a_t(x)=0$ and $a(t)=k$ for k constant ? — Carlos Andres Henao Acevedo, Apr 13 '18 at 22:50
@CarlosAndresHenaoAcevedo Actually if a question is asked without showing efforts to solve it, it will probably get down votes and some users will also maliciously downvote answers to such a question. In this case, my answer is correct but it got down vote. — Ѕᴀᴀᴅ, Apr 14 '18 at 05:13

score 2 · Accepted Answer · answered Apr 14 '18 at 03:18

$\def\d{\mathrm{d}}$Case 1: $c^2 \neq 1$. Make substitution $(y, s) = (x - c^2 t, t - x)$, then$$ u_t = u_s - c^2 u_y, \quad u_{tt} = u_{ss} - 2c^2 u_{ys} + c^4 u_{yy}, \quad u_{xx} = u_{yy} - 2u_{ys} + u_{ss}, $$ and the equations become$$ \begin{cases} (1 - c^2) u_{ss} - (c^2 - c^4) u_{yy} = 0, \quad y \in \mathbb{R},\ s > 0\\ u\bigr|_{s = 0} = φ\left( \dfrac{y}{1 - c^2} \right), \quad y \in \mathbb{R}\\ (u_s - c^2 u_y)\bigr|_{s = 0} = 0. \quad y \in \mathbb{R} \end{cases} \tag{1} $$ Because $u\bigr|_{s = 0} = φ\left( \dfrac{y}{1 - c^2} \right)$, then$$ u_y\bigr|_{s = 0} = \frac{\d}{\d y} \left( φ\left( \frac{y}{1 - c^2} \right) \right) = \frac{1}{1 - c^2} φ'\left( \frac{y}{1 - c^2} \right), $$ and (1) becomes$$ \begin{cases} u_{ss} - c^2 u_{yy} = 0, \quad y \in \mathbb{R},\ s > 0\\ u\bigr|_{s = 0} = φ\left( \dfrac{y}{1 - c^2} \right), \quad y \in \mathbb{R}\\ u_s\bigr|_{s = 0} = \dfrac{c^2}{1 - c^2} φ'\left( \dfrac{y}{1 - c^2} \right). \quad y \in \mathbb{R} \end{cases} \tag{2} $$ Thus\begin{align*} u(y, s) &= \frac{1}{2} \left( φ\left( \frac{y + cs}{1 - c^2} \right) + φ\left( \frac{y - cs}{1 - c^2} \right) \right) + \frac{1}{2c} \int_{y - cs}^{y + cs} \dfrac{c^2}{1 - c^2} φ'\left( \dfrac{ξ}{1 - c^2} \right) \,\d ξ\\ &= \frac{1}{2} \left( φ\left( \frac{y + cs}{1 - c^2} \right) + φ\left( \frac{y - cs}{1 - c^2} \right) \right) + \frac{c}{2} \left( φ\left( \frac{y + cs}{1 - c^2} \right) - φ\left( \frac{y - cs}{1 - c^2} \right) \right)\\ &= \frac{1 + c}{2} φ\left( \frac{y + cs}{1 - c^2} \right) + \frac{1 - c}{2} φ\left( \frac{y - cs}{1 - c^2} \right), \end{align*} and$$ u(x, t) = \frac{1 + c}{2} φ\left( \frac{x + ct}{1 + c} \right) + \frac{1 - c}{2} φ\left( \frac{x - ct}{1 - c} \right). $$

Case 2: $c^2 = 1$. Make substitution $(y, s) = (x, t - x)$, then$$ u_t = u_s, \quad u_{tt} = u_{ss}, \quad u_{xx} = u_{yy} - 2u_{ys} + u_{ss}, $$ and the equations become$$ \begin{cases} 2u_{ys} - u_{yy} = 0, \quad y \in \mathbb{R},\ s > 0\\ u\bigr|_{s = 0} = φ(y), \quad y \in \mathbb{R}\\ u_s\bigr|_{s = 0} = 0, \quad y \in \mathbb{R} \end{cases} \tag{3} $$ which implies$$ \begin{cases} 2u_s - u_y = η(s), \quad y \in \mathbb{R},\ s > 0\\ u\bigr|_{s = 0} = φ(y), \quad y \in \mathbb{R} \end{cases} \tag{4} $$ where $η(s)$ is a function to be determined. By the method of characteristics, the general solution to (4) is$$ u(y, s) = \frac{1}{2} \int_0^s η(ξ) \,\d ξ + φ\left( y + \frac{s}{2} \right), $$ and plugging it back to (3) yields$$ 0 = u_s\bigr|_{s = 0} = \frac{1}{2} η(s) + \frac{1}{2} φ(y) = 0. \quad \forall y \in \mathbb{R} $$ For the existence of solutions to the original equations, $φ$ must be a constant (namely $-η(0)$) and$$ u(x, t) = \frac{1}{2} \int_0^{t - x} η(ξ) \,\d ξ - \frac{1}{2} η(0). $$ It can be verified that $u(x, t)$ with the form above are indeed solutions.

thanks, in the case 1, why $\quad u_{xx} = u_{yy} - 2u_{ys} + u_{ss}$? I have $\quad u_{xx} = 2u_y+4x^2u_{yy} - 4xu_{ys} + u_{ss}$ — Carlos Andres Henao Acevedo, Apr 14 '18 at 13:02
@CarlosAndresHenaoAcevedo Because $\partial_x=\partial_y-\partial_s$, then $\partial_{xx}=(\partial_y-\partial_s)^2=\partial_{yy}-2\partial_{ys}+\partial_{ss}$. — Ѕᴀᴀᴅ, Apr 14 '18 at 13:14

How solve the following PDE?

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