# I Separation of Variables for Partial Differential Equations

#### FAS1998

When using the separation of variable for partial differential equations, we assume the solution takes the form u(x,t) = v(x)*g(t).

What is the justification for this?

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#### Dr Transport

Science Advisor
Gold Member
it works.....and it can be shown that the solutions are mathematically rigorous.

• FAS1998

#### wrobel

in PDE "separation of variable" is just an archaic and misleading look. Actually we expand a solution in a basis of eigenfunctions of linear differential operator that is related to equation

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• FAS1998

#### FAS1998

it works.....and it can be shown that the solutions are mathematically rigorous.
I understand that it works in the sense that the solutions it finds are consistent with the differential equations, but how do we know that the solutions couldn’t be more general?

Couldn’t solutions of u(x,t) exist that are not of the form v(x)*g(t)?

Or can all functions of u(x,t) be written as v(x)*g(t)?

#### Dr Transport

Science Advisor
Gold Member
I understand that it works in the sense that the solutions it finds are consistent with the differential equations, but how do we know that the solutions couldn’t be more general?

Couldn’t solutions of u(x,t) exist that are not of the form v(x)*g(t)?

Or can all functions of u(x,t) be written as v(x)*g(t)?
For linear equations, you can write them in the form $u(x,t) = v(x)g(t)$. For non-linear equations that may not be the form of the solution.

• FAS1998

#### FAS1998

For linear equations, you can write them in the form $u(x,t) = v(x)g(t)$. For non-linear equations that may not be the form of the solution.
So the solution of all linear PDE’s take the form u(x,t) = v(x)g(t)?

Is there a proof or explanation for this somewhere? My textbook doesn’t explain this very thoroughly.

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#### Dr Transport

Science Advisor
Gold Member
see wrobel's post above....

#### bigfooted

Most books that I have on this topic do not give any explanation. An exception is Morse and Feshback, Methods of Theoretical Physics, part I.
Later, it was found that (Lie) symmetries play a large role in determining separable coordinate systems, Willard Miller has written a book about it (which is still on my to-read list).
There is some 'further reading' here:

#### phyzguy

Science Advisor
So the solution of all linear PDE’s take the form u(x,t) = v(x)g(t)?
No, certainly not.

The idea is that you try to find a solution in this form. If it works, you've found a solution. If it doesn't, you need to try other methods. Usually it doesn't work, but it's worth a try. Here's a similar situation. Say I'm trying to find the square root of 169. It's obviously greater than 10, and less than 20. So I try 11x11=121, 12x12=144, 13x13=169... hey guess what? 13 works! So I've found the solution through a simple method. If I'm trying to find the square root of 170, obviously this method won't work, so i need to try more advanced methods

• FAS1998

#### WWGD

Science Advisor
Gold Member
So the solution of all linear PDE’s take the form u(x,t) = v(x)g(t)?

Is there a proof or explanation for this somewhere? My textbook doesn’t explain this very thoroughly.
Maybe this is contrived, but if u(x,t) is the pdf of a joint distribution where x,t, are not independent, then the equation will not separate.

#### robphy

Science Advisor
Homework Helper
Gold Member
So the solution of all linear PDE’s take the form u(x,t) = v(x)g(t)?

Is there a proof or explanation for this somewhere? My textbook doesn’t explain this very thoroughly.
One you have that form u(x,t) = v(x)g(t) and you found the forms of v(x) and g(t) from their ODEs (and possibly some of the boundary and initial conditions),
you then exploit the linearity of the PDE to write the general solution as a linear combination of the v(x)g(t)'s.

#### FAS1998

No, certainly not.

The idea is that you try to find a solution in this form. If it works, you've found a solution. If it doesn't, you need to try other methods. Usually it doesn't work, but it's worth a try. Here's a similar situation. Say I'm trying to find the square root of 169. It's obviously greater than 10, and less than 20. So I try 11x11=121, 12x12=144, 13x13=169... hey guess what? 13 works! So I've found the solution through a simple method. If I'm trying to find the square root of 170, obviously this method won't work, so i need to try more advanced methods
How do you know that there isn’t a more general solution? Couldn’t it be possible that some solution of the form v(x)g(t) exists, but is not the most general form?

Or in other words, how do you know if sepetation or variables has worked, and has provided a general solution?

#### phyzguy

Science Advisor
How do you know that there isn’t a more general solution? Couldn’t it be possible that some solution of the form v(x)g(t) exists, but is not the most general form?

Or in other words, how do you know if sepetation or variables has worked, and has provided a general solution?
For most PDEs where you are applying the method of separation of variables, there is a proof of uniqueness. So it you have found a solution, you have found the solution.

• FAS1998

#### jordi

I was thinking about this also. An answer is the one they have given to you: for some PDE (with some boundary conditions), the solution is unique. For example, see Zeidler volume 108.

As a consequence, if a naive separation of variables work, then you expand a general solution with this "basis", and if you can find coefficients in this general solution that satisfies your boundary conditions, that is the solution.

I have looked into the literature, and I have been unable to find general statements, by reading some new books on PDE. For example, it would be "natural" that for some type of PDE, one could guarantee that the separation of variables "basis" generates all solutions of the PDE. But it seems that it is not easy (or even possible) to do this kind of statements.

#### haushofer

Science Advisor
Of course, the variable-separated solution looks way too constrained. But then you use linearity of the PDE and it turns out that often these solutions form a complete basis of your solution space! Hence any solution can be written in this basis. It is just a matter of finding the right coefficients. And from linear algebra you know how to calculate those.

If it works, that is. E. g. if you take the Schrodinger eqn. with time-dependant potential, the party is spoiled.

#### Ssnow

Gold Member
$u(x,t)\,=\, v(x)\cdot g(t)$ is also a "sort" of factorization assumption ...
Ssnow

#### pasmith

Homework Helper
How do you know that there isn’t a more general solution? Couldn’t it be possible that some solution of the form v(x)g(t) exists, but is not the most general form?

Or in other words, how do you know if sepetation or variables has worked, and has provided a general solution?
Uniqueness theorems: If you find a solution, by whatever ad hoc non-a-priori-justified method, then it must be the solution.

• haushofer

#### HallsofIvy

Science Advisor
Homework Helper
You keep writing as if you think "u(x)v(t)" is all there is. No one has said that! What is the situation is that the general solution can be written as linear combinations of all the different "u(x)v(t)" forms. In applications that leads to generalized "Fourier series" (over finite intervals) and "Fourier transforms" (over infinite intervals).

#### mpresic3

When using the separation of variable for partial differential equations, we assume the solution takes the form u(x,t) = v(x)*g(t).

What is the justification for this?

This bothered me when I was an undergraduate studying separation of variables for partial differential equations. About a month ago, a much younger co-worker and college asked me to justify why we can calculate the gravitational field with partial differential equation. He is in the middle of his undergraduate education and the only technique he reported knowing for computing the field was direct integration using mass points and Newtons law or Gausses law (if he applied it to gravitation). His actual question involved why a collection of point masses each contributing an inverse square law could ever generate an inverse cube law in the field.
Luckily of both of us, there was a interruption that gave both of us some time. I allowed myself the luxury of thinking long and hard about the question offered. When we came back to the office, after I brought up the point dipole, he eagerly demonstrated he knew how to compute the electrical potential (and later the field).
Clearly the electric dipole had generated an inverse cube term in the field. I alluded to the similarity with gravity and he was happy with this. (I later instructed him that the relationship was not entirely similar as the weakening of the field is due also to the screening with unlike charges, but the point was made.)

I find most books and educators concentrate on "proofs" and applications and not on justifications. The example above does not "prove" anything but it justifies analysis with differential equations as an alternative to integrals. (Actually Gausses law can also be used in differential form.)

After this long winded prelude, I found in a similar manner, the potential due to a square quadrupole of electric charge lying in a plane. The answer is "something like" (I do not have my notes in front of me)

V = ( A / r to the cube ) * cosine ( theta ) * sin ( 2. phi ).

I apologize with my notes gone this is probably wrong but the salient points are valid. The solution for the potential is a function of r times a function of theta times a function of phi.

Now back to the textbook. One example does not "prove" anything, but wouldn't this example placed before the section (presentation) of separation of variables " suggest " or justify or at least motivate this ansatz.

I think a textbook demonstrating justifications and motivations for physics techniques would be well received if it were written. I think these are very rare. Maybe other readers can follow on

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