(adsbygoogle = window.adsbygoogle || []).push({}); Question about Solutions of second order linear PDEs

I don't have very much formal knowledge of this topic, this is something I have been thinking about, so excuse me if my notation is off. I have a question about second order linear PDEs, do all have a separable solution? It seems that we can create a general separated system of operators that will mesh together to form any given second order linear equation.

Assume that [itex]u(x,t,y,...)=f_{1}(x)f_{2}(t)f_{3}(y)...[/itex]

[itex](D_{xx}+a_{1}D_{x}+a_{2})(u)=0[/itex]

[itex](D_{tt}+b_{1}D_{t}+b_{2})(u)=0[/itex]

[itex](D_{yy}+c_{1}D_{y}+c_{2})(u)=0[/itex]

so on and so forth. Then we solve each resulting single variable equations.

[itex]f''_{1}(x)+a_{1}f'_{1}(x)+a_{2}f_{1}(x)=0[/itex]

[itex]f''_{2}(t)+b_{1}f'_{2}(t)+b_{2}f_{2}(t)=0[/itex]

[itex]f''_{3}(y)+c_{1}f'_{2}(y)+c_{2}f_{3}(y)=0[/itex]

So here is one example...

[itex]u_{tt}=u_{xx}+u_{yy}[/itex]

I work backwards, assume [itex]u(t,x,y)=G(t)f_{1}(x)f_{2}(y)[/itex]

[itex]G(t)f''_{1}(x)f_{2}(y)+G(t)f_{1}(x)f''_{2}(y)=G''(t)f_{1}(x)f_{2}(y)[/itex]

I use two constants to separate [itex]\lambda , v[/itex]

And work it out to three equations...

[itex]f''_{1}(x)+vf_{1}(x)=0[/itex]

[itex]f''_{2}(y)+(\lambda-v)f_{2}(y)=0[/itex]

[itex]G''(t)+\lambda G(t)=0[/itex]

So, my question is, did I come to the right assumption in that all second order linear equations have a separable solution?

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# Solutions of second order linear PDEs

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