Solutions of second order linear PDEs

[saminator910]

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 u(x,t,y,...)=f_{1}(x)f_{2}(t)f_{3}(y)...

(D_{xx}+a_{1}D_{x}+a_{2})(u)=0

(D_{tt}+b_{1}D_{t}+b_{2})(u)=0

(D_{yy}+c_{1}D_{y}+c_{2})(u)=0

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

f''_{1}(x)+a_{1}f'_{1}(x)+a_{2}f_{1}(x)=0

f''_{2}(t)+b_{1}f'_{2}(t)+b_{2}f_{2}(t)=0

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

So here is one example...

u_{tt}=u_{xx}+u_{yy}

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

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

I use two constants to separate \lambda , v

And work it out to three equations...

f''_{1}(x)+vf_{1}(x)=0
f''_{2}(y)+(\lambda-v)f_{2}(y)=0
G''(t)+\lambda G(t)=0

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

 

[pasmith]

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?

No. Whether a PDE has a separable solution doesn't just depend on the PDE; it also depends on the boundary of the region in which the PDE is to be solved.

A practical criterion is that the boundary must consist of coordinate surfaces in a coordinate system in which the PDE is separable.

For example, Laplace's equation \nabla^2 u = 0 does not have a separable solution if it is to be solved in a region A \subset \mathbb{R}^2 whose boundary is an arbitrary simple closed curve. Only if the boundary consists of coordinate surfaces in one of the systems listed here is there a separable solution.

 

[saminator910]

Ok, so what you are saying is that the coordinate system which the PDE is defined on will also effect whether it is separable. So, where the wave equation may be separable in Cartesian coordinates, it may not in Spherical coordinates?

 

[AlephZero]

I would put what pasmith was saying a slightly different way. Of course your PDE has an infinite number of separable solutions, because you just found them in terms of two arbitrary parameters $\lambda$ and $\nu$.

But if you want to find a solution with a given set of boundary conditions, somehow you have to find equations for $\lambda$ and $\nu$ in terms of those boundary conditions.

You can only do that easily if the form of your solutions has a simple relationship to the shape of the boundary.

Often, to satisfy some part of the boundary conditions means that $\lambda$ etc can only take a discrete set of values $\lambda_i, i = 0, 1, 2, \dots$, so your can write a general solution as the sum of an infinite series of functions, similar to a Fourier series. The unknown coefficients in the series are found by using the rest of the boundary conditions.

A PDE like the the wave equation will be separable in almost any "useful" coordinate system, but you need to use a coordinate system that matches the geometry of the problem and the symmetry that you expect in the solution - for example rectangular, cylindrical, spherical, elliptic-hyperbolic, or something even more specialized.

The form of the solutions you get will depend what coordinate system you use. If you separate the variables for the wave equation in rectangular coordinates, you will get solutions that look like plane waves, not spherical waves. In principle you could define a spherical wave as an infinite sum of plane waves, but it's much easier to sidestep that problem by using spherical coordinates.

 

[saminator910]

Oh, alright, let me get this straight. The equation itself will always be separable, but that does not necessarily imply that it will have a solution on any given boundary conditions. For that we must have the right combination of coordinate system and boundary?

But I will be able to separate out any given second order linear PDE into a resulting system of single variable equations, correct? But again, that doesn't necessarily imply a solution.