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This feels like a question that is both simple and duplicate but I can't find an answer or a previous version of the question.

Suppose we are given some PDE, for example Laplace's Equation in polar coordinates, $$ \nabla^2 u = \frac{\partial^2u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2u}{\partial \theta^2} = 0. $$ The usual way to solve this kind of equation is to posit a separated solution $$u(r,\theta) = R(r)\Theta(\theta).$$ We then get something like $$(R'' + \frac{1}{r} R') \Theta = -\frac{R}{r}\Theta''.$$ At this point we usually divide by $$\frac{\Theta R}{r}$$ to find two ODEs.

Now, I can see that $\Theta R$ can't be the zero function (well, maybe it can be, but we don't like them zero solutions). But what if $\Theta R$ is zero at some specific point? Are we looking for solutions that are everywhere non-zero?

For example, the exercise that I am working on right now is the above problem on the half-sphere $$ H = \{ (x,y)\ | \ x^2 +y^2 \leq a, y \geq 0 \},$$ with boundary conditions $$u(a,\theta) = \sin \theta,$$ $$ u(r,0) = u(r,\pi) = 0.$$

To summarize: I know how to solve the problem, I just don't know how to justify the dividing by $\Theta R$. I can think of one argument: we can try if the answer we get is a solution by substituting in the original equation later. But then it is not clear in general if we have lost any solutions?

Thanks for any help in advance.

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  • $\begingroup$ As you said, separation of variables is a method to find the explicit form of the solution. The justification is "it works". The uniqueness should be obtained later (by a standard contradiction argument, for instance). Furthermore, the possibility of losing some solution appears at the same time we make the ansatz of sepration of variables $$ u(x,y)=X(x)Y(y). $$ $\endgroup$ – guacho Apr 2 '16 at 22:16
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At the point "We get something like $\ldots$" we need the following Lemma, written out in terms of variables $x$ and $y$ instead of $r$ and $\theta$:

Lemma. Let $I$ and $J$ be intervals on the $x$-, resp., the $y$-axis. Assume that $$a(x)b(y)=c(x)d(y)\qquad\forall x\in I,\quad\forall y\in J\tag{1}$$ with both sides $\ne0$ for at least one point $(x_0,y_0)$. Then there is a constant $\mu\ne0$ with $$c(x)=\mu a(x)\quad(x\in I),\qquad b(y)=\mu d(y)\quad(y\in J)\ .$$ Proof. Put $x:=x_0$ in $(1)$ and obtain $$b(y)={c(x_0)d(y)\over a(x_0)}\quad(y\in J)$$ and therefore $b(y)=\mu d(y)$ $(y\in J)$ with $\mu={\displaystyle{c(x_0)\over a(x_0)}}\ne0$. Now put $y:=y_0$ in $(1)$ and obtain $$a(x)\>\mu d(y_0)=c(x) d(y_0)\quad(x\in I)\ .$$ This implies $c(x)=\mu a(x)$ $(x\in I)$, as claimed.

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  • $\begingroup$ Can you explain in more detail how this solves the problem? $\endgroup$ – Timon van der Berg Apr 13 '16 at 18:05

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