Let $g$ be a twice differentiable function, set $f = g''(x)$ and assume that both $\displaystyle\int_\mathbb{R} |f(x)| \mathrm{d}x<\infty$ and $\displaystyle\int_\mathbb{R} |g(x)| \mathrm{d}x<\infty$ such that the fouriertransforms of $f$ and $g$ exists. We want to find a solution of the differential equation $ \frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial y} = 0 \qquad \quad -\infty < x < \infty, \quad y \geq 0$ such that $ \lim_{x \to \pm \infty} u(x,y) = \lim_{x\to\pm\infty} \frac{\partial u}{\partial x}(x,y)=0 $ and finally $\frac{\partial u}{\partial x}(x,0)= f(x)$.
I am trying to prepare for my exam, and this was one of the questions. It asks me to transform the problem by using the Fourier transform into a ordinary differential equation and solve it.
But I am a bit stuck, there is so much information, and I have never done a Fourier transform before, could anyone be so kind as to help me? =)
I assumed that the solution could be written on the form $u(x,y)=X(x)Y(y)$ such that I end up with
$ \ddot{X} Y = Y' X \ \Rightarrow \ \frac{\ddot{X}}{X} = \frac{Y'}{Y}$
Which can be assumed to be equal to a complex constant $\gamma^2$. Now this gives the equations
$ \ddot{X} = \gamma^2 X \\ Y' = \gamma^2 Y $
Solving the top equation gives me a solution on the form
$ X(x) = A e^{i \lambda x} + B e {-i \lambda x} $
But I am having problems seeing how this solution can satisfy the condition that
$ \lim_{x \to \pm \infty} u(x,y) = Y(y) \lim_{x \to \pm \infty} X(x) = 0$
Now, I can not really see how I can pick $A$, $B$ and $\gamma$ to satisfy this. Any suggestions?