# Solve $U_{tt}=a U_{xx}$ when a<0.

I know $U_{tt}=a U_{xx}$ when $a>0$ is a wave equation, and we can solve with d'Alembert's formula.

How about solving $U_{tt}= a U_{xx}$ when $a<0$ within $x\in[0,1]$ with initial condition u(x,0)=1.

Where can I start with this PDE? Any hints or theorem on this? Thank you!

• Have you tried the method that works for $a > 0$? What issues have you run into? – Mark Jan 16 '18 at 3:08
• Will $a<0$ still be a wave equation? – Ying Jan 16 '18 at 3:12
• $U_{tt} +|a|U_{xx} = 0$ is a Laplace equation in the two variables. All that is needed is to rescale in $x$ to turn it into a Laplace equation. – DisintegratingByParts Jan 16 '18 at 5:14

$$U_{tt}= a U_{xx} \quad\text{when}\quad a<0 , \quad\text{with initial condition}\quad U(x,0)=1 \tag 1$$ The specified condition is not sufficient. They are an infinity of solutions as show below.

Since $a<0$, let $a=-b^2$.

$$U_{tt}= -b^2 U_{xx}$$

Let $U(x,t)=V(x,t)+1$ $$V_{tt}= -b^2 V_{xx} \quad\text{with initial condition}\quad V(x,0)=0 \tag 2$$

A general form of solution is : $$V(x,t)=f(x+ibt)+g(x-ibt) \tag 3$$ where $f$ and $g$ are any differentiable functions. Just put it into Eq.$(2)$ and see that $3$ is convenient.

Condition :

With any $X\::\quad V(X,0)=0=f(X)+g(X) \quad\implies\quad g(X)=-f(X)$

$V(x,t)=f(x+ibt)-f(x-ibt)\quad\to\quad U(x,t)=f(x+ibt)-f(x-ibt)+1$ $$U(x,t)=f(x+i\sqrt{-a}\:t)-f(x-i\sqrt{-a}\:t)+1$$