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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:

enter image description here

Does this seem reasonable?

Thanks

Luthier415Hz
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  • What are the initial conditions for $u$, exactly? That's the wave equation, which you usually solve by using $x \pm y$ as new variables, rather than with separation of variables. – Hans Lundmark Dec 09 '21 at 18:16
  • I take they have to be multiplied as $u(x,y)=f(x)*g(y)$ – Luthier415Hz Dec 10 '21 at 08:57
  • You didn't answer my question. You give conditions for $f$ and $g$, but where did you get them from? What was the formulation of original problem (and in what context did you encounter it)? What were the conditions on the function $u$? – Hans Lundmark Dec 10 '21 at 09:32
  • They come from an exercise that was hand-written by a Professor, that is, they do not come from a printed book. So to me, it appears that he has done a mistake on the double-valuedness of f(x) in the IC. – Luthier415Hz Dec 10 '21 at 09:50

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