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I'm having trouble understanding the separation of variables technique for PDEs. I know it has something to do with assuming the solution u(x,y) is in the form X(x)Y(y)? But I've never learned Fourier series before and I was having trouble following along in class.

For example, take this (seemingly) basic PDE. The examples we did in class were much harder and I think I need to understand it first using a much more rudimentary one.

$u_x + u_y = 0$

How would one go about using the separation of variables technique to find solutions u(x,y) for this PDE? I'd really appreciate a step by step explanation of how to go about solving this. Thanks in advance.

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    $\begingroup$ See here for an example that I worked out math.stackexchange.com/questions/671085/… $\endgroup$
    – rajb245
    Commented Feb 10, 2014 at 18:22
  • $\begingroup$ Also, you have to give boundary conditions to have a well-specified problem that someone could work out for you. $\endgroup$
    – rajb245
    Commented Feb 10, 2014 at 18:23
  • $\begingroup$ My prof gave us this as an easy example to work on out our own but he didn't give us any boundary conditions... $\endgroup$
    – user114014
    Commented Feb 10, 2014 at 18:24
  • $\begingroup$ You can also carefully read through en.wikipedia.org/wiki/…, look at the picture to make sense of the boundary conditions. $\endgroup$ Commented Feb 10, 2014 at 18:35

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Let $u(x,y)=X(x)Y(y)$, then plug that in to your original equations $$ X'(x)Y(y) + X(x)Y'(y) = 0 $$ Dividing through by $X(x)Y(y)$ and rearranging $$ \frac{X'(x)}{X(x)} = -\frac{Y'(y)}{Y(y)} $$ The left hand side depends only on the variable $x$, and the right hand side depends only on the variable $y$, yet they are equal to each other for all $(x,y)$. Each side must therefore be a constant (call it $\lambda$), giving two equations $$ \frac{X'(x)}{X(x)} = \lambda \\ \frac{Y'(y)}{Y(y)} = -\lambda $$ Rearranging gives $$ X'(x) - \lambda X(x) = 0 \\ Y'(y) + \lambda Y(y) = 0 $$ These are first order ODEs you can solve. The solutions are $$ X(x) = e^{\lambda x} \\ Y(y) = e^{-\lambda y} $$ A general solution (for any $\lambda$) is $$ u_\lambda(x,y) = c(\lambda)e^{\lambda x}e^{-\lambda y} = c(\lambda)e^{\lambda (x-y)} $$ Without boundary conditions, this is as far as you can go. The boundary conditions would help you pin down what $\lambda$ and $c(\lambda)$ are. For an equation with at least two derivatives, that's also when the Fourier series side of it comes into play. Usually you get a discrete spectrum of $\lambda$ values, and the solution is a summation of terms like that over the different values of $\lambda$. Problems on unbounded domains can have a continuous spectrum of $\lambda$ values and you end up with an integral instead of a summation (a Fourier transform instead of a Fourier series essentially).

For this problem, with the right interpretation of the terms, the general solution will always be $$ u(x,y) = \int_{-\infty}^\infty c(\lambda)e^{\lambda (x-y)}\;d\lambda $$ which can reduce to a discrete summation with a bit of help from complex analysis.

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  • $\begingroup$ Thanks so much - all of this makes sense to me. Another example he gave us was u_x + u_y = (x + y)u. In this case, I understand what to do up to when you divide by X(x)Y(y) and get X'(x)/X(x) + Y'(y)/Y(y) = x+y. How should you handle it if it's not equal to 0? $\endgroup$
    – user114014
    Commented Feb 10, 2014 at 18:54
  • $\begingroup$ It still separates out into parts that depend on $x$ only, and on $y$ only. In that case you could rearrange to $X'(x)/X(x)-x=\lambda=-Y'(y)/Y(y)+y$. $\endgroup$
    – rajb245
    Commented Feb 10, 2014 at 19:27

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