Treating one of the variables in Laplace equation as time

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I'm a bit stuck on how to approach this question.

Suppose someone treats one of the variables in a Laplace equation as time and tries to solve the evolution problem,

$$u_{tt}+u_{xx}=0\ (0<x<l\,,\ t>0)\,,\quad u(0,t)=u(l,t)=0\,,\quad u(x,0)=\phi(x)\,,\quad u_t(x,0)=\psi(x) $$

similar to the wave equation. Show that this problem has no continuous dependence on data, even if the time $t$ belongs to a finite interval $0<t<T$ ($T$ is a positive constant).

I initially thought trying to solve with separation of variables so you have,

$$\frac{\ddot{T}(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda$$

So you end up with ,

$$T(t) = \sin(\sqrt{\lambda}t) + \cos(\sqrt{\lambda} t)$$ and $$X(x) = A e^{\sqrt{\lambda x}} + B e^{-\sqrt{\lambda x}}$$

and using the initial conditions on $X$ you end up with,

$$X(0) = A+B=0\implies B=-A$$

$$X(l) = A(e^{\sqrt{\lambda l}} - e^{-\sqrt{\lambda l}}) = 0$$

and since we want a non-trivial solution we solve for $\lambda$ and get that $\lambda = 0$ or imaginary, but I'm not sure if imaginary eigenvalues are allowed for this type of question.

Is this the correct way to approach it or should I try to use d'Alembert's formula where $c=i$?

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I was going about it in the wrong way. Because the goal is just to show the instability it's enough to say that there is a solution

$$u(x,t)=A\sin(n x)e^{n t}$$

since

$$u_{tt} = -A n^2\sin(n x)e^{n t}\,,\quad u_{xx} = A n^2 \sin(n x)e^{n t}$$

and we can set $l=\pi$ for simplicity

Then if we let $A=\frac{1}{n}$

$$u(x,0)\rightarrow_{n\rightarrow \infty} 0$$ $$u(x,1)\rightarrow_{n\rightarrow\infty} \infty$$

which shows that since it is ill-posed with respect to the parameter $n$ so it has no continuous dependence on the data.