I have the given PDE:
$u_{xx}=u_{yy}$
where the solution must be in the form: $u(x,y)=f(x)\cdot g(y)$.
By solving this using the product rule, I get
\begin{equation} \begin{array} \\ f_{xx}=f(x)C\\ g_{yy}=g(y)C\\ \end{array} \end{equation}
which I solve separately, and obtain
\begin{equation} f(x)=e^{\sqrt{c}x}+e^{-\sqrt{c}x} \end{equation}
\begin{equation} g(y)=e^{\sqrt{c}y}+e^{-\sqrt{c}y} \end{equation}
Since $u(x,y)=f(x)\cdot g(y)$ I multiple the solutions and obtain:
\begin{equation} (e^{\sqrt{c}x}+e^{-\sqrt{c}x})(e^{\sqrt{c}y}+e^{-\sqrt{c}y}) = 4 cosh(\sqrt{c}x) cosh(\sqrt{c}y) \end{equation}
Then they have to satisfy the I.C (which appear double-valued):
\begin{equation} \begin{array} ff(0)=0, \ f(1)=0, \ f(1/4)=1/2, \\ g(1/4)=g(1/2)=1 \end{array} \end{equation}
But since the I.C. are double-valued, which appears wrong, I change I.C. to $f(0.25)=g(0.25)=1$
So I do: \begin{equation} 4 cosh(\sqrt{c}\cdot0.25) cosh(\sqrt{c}\cdot0.25)=1 \end{equation}
and I obtain $c=-\frac{16\pi^2}{9}$. So, the full solution appears to be:
\begin{equation} u(x,y)=4 cosh(\frac{4i\pi}{3}x) cosh(\frac{4i\pi}{3}y) \end{equation}
The solution is then plotted:
Does this seem reasonable?
Thanks
