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Can anybody explain me why the method of separation of variables for linear homogeneous PDE works ? thanks

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If you are looking for a proof: – emka Dec 26 '12 at 17:27
That page is ODE, not PDE, where "separation of variables" has a different meaning. – GEdgar Dec 26 '12 at 17:50
What do you mean by "works"? Do you mean, why does the method give a solution, or why does the method give,say, a unique solution? – Tom Oldfield Dec 26 '12 at 19:10
@TomOldfield Exactly – Real Hilbert Dec 27 '12 at 6:15
@RealHilbert I'm confused, I asked an "or" question, what do you mean by "exactly"? – Tom Oldfield Dec 27 '12 at 17:36

"Separation of variables" has two totally different meanings, which have nothing to do with each other.

(i) A general ODE is of the form $y'=f(x,y)$, where $f$ is an arbitrary function of two variables $x$, $y$. Sometimes this function is not so arbitrary, but is a product of two functions depending only on $x$, resp., on $y$: $$y'=g(x) h(y)\ .$$ In this case writing formally $y'={dy\over dx}$ you can "separate the variables $x$ and $y$" and obtain the equation $${1\over h(y)}\ dy\ =\ g(x)\ dx\ .$$ There is a certain formal procedure of integration on the left with respect to $y$ and on the right with respect to $x$ that finally produces all solutions of the given ODE. The proof that this works is not at all intuitive.

(ii) Given a linear homogeneous PDE in several variables, say: two variables $x$ and $t$ that together range over a maybe infinite rectangle in the $(x,t)$-plane, you can raise the question whether there are "special solutions" $f$ of the form $f(x,t)=X(x)\cdot T(t)$. For $f$ to be such a special solution it will be necessary that $X(\cdot)$ and $T(\cdot)$ satisfy certain linear ODEs which are easy to solve. When you are lucky you get a large bag full of such special solutions $f_\lambda(x,t)=X_\lambda(x)T_\lambda(t)$, $\ \lambda\in\Lambda$.

Now it is a fundamental feature of linear homogeneous systems of any kind that any linear combination of known solutions is again a solution. It follows that any function $u$ of the form $$u(x,t)=\sum_{\lambda\in\Lambda} c_\lambda X_\lambda(x)T_\lambda(t)$$ is a solution of your PDE. Therefore it makes sense to try to determine the coefficients $c_\lambda$ in such a way that the other conditions (in particular, boundary conditions or initial conditions) present in your problem are met as well.

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so these special solutions always there for all linear homogeneous PDE, Are they only special solutions for a linear homogeneous PDE? – Real Hilbert Dec 27 '12 at 6:26
@Real Hilbert: A function $f$ of two independent variables $x$ and $t$ that is of the form $f(x,t)=X(x)\cdot T(t)$ is always "special". – Christian Blatter Dec 27 '12 at 9:48

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