Method of separation of variables

[evinda]

Hello! (Wave)

I want to check if the method of separation of variables can be used for the replacement of the following given partial differential equations from a pair of ordinary differential equations. If so, I want to find the equations.- $[p(x) u_x]_x-r(x) u_{tt}=0$
- $u_{xx}+(x+y) u_{yy}=0$
- $u_{xx}+u_{yy}+ xu=0$For the first pde, I thought the following:

Suppose that $u(t,x)=X(x) T(t)$. Then $[p(x) u_x]_x=p'(x) u_x+ p(x) u_{xx}=p'(x) X'(x) T(t)+ p(x) X''(x) T(t)=T(t) [p'(x) X'(x)+ p(x) X''(x)]$

So $$T(t) [p'(x) X'(x)+ p(x) X''(x)]=r(x) X(x) T''(t) \Rightarrow \frac{[p'(x) X'(x)+ p(x) X''(x)]}{r(X) X(x)}=\frac{T''(t)}{T(t)}=-\lambda$$

So we get the following system of ordinary differential equations:

$\left\{\begin{matrix}
p(x) X''(x)+p'(x) X'(x)+ \lambda r(x) X(x)=0\\
T''(t)+\lambda T(t)=0
\end{matrix}\right.$

Can we find somehow the general form of the solution of this DE:

$p(x) X''(x)+p'(x) X'(x)+ \lambda r(x) X(x)=0$ ?

Also by solving the system, we will have found one solution of the pde, we will not know if the solution will be unique, right? Would the same happen if we would also have two boundary consitions as for $x$ and an initial condition as for $t$?

For the second pde:

Suppose that $u(x,y)= X(x) Y(y)$. Then $u_{xx}+(x+y) u_{yy}=0 \Rightarrow X''(x) Y(y)+ (x+y) X(x) Y''(y)=0$.

So this cannot be solved with the method of separation of variables. And so we deduce that the solution isn't of the form $X(x) Y(y)$, right? How can we justify it more formally that this method cannot be used?For the last pde, I found the following system: $\left\{\begin{matrix}
X''+(x-\lambda)X=0\\
Y''+ \lambda Y=0
\end{matrix}\right.$

We don't know a genaral methodology to solve the first DE of the latter system, do we?

 

[HallsofIvy]

Re: Method of separation of variables

Hello! (Wave)

I want to check if the method of separation of variables can be used for the replacement of the following given partial differential equations from a pair of ordinary differential equations. If so, I want to find the equations.- $[p(x) u_x]_x-r(x) u_{tt}=0$
- $u_{xx}+(x+y) u_{yy}=0$
- $u_{xx}+u_{yy}+ xu=0$For the first pde, I thought the following:

Suppose that $u(t,x)=X(x) T(t)$. Then $[p(x) u_x]_x=p'(x) u_x+ p(x) u_{xx}=p'(x) X'(x) T(t)+ p(x) X''(x) T(t)=T(t) [p'(x) X'(x)+ p(x) X''(x)]$

So $$T(t) [p'(x) X'(x)+ p(x) X''(x)]=r(x) X(x) T''(t) \Rightarrow \frac{[p'(x) X'(x)+ p(x) X''(x)]}{r(X) X(x)}=\frac{T''(t)}{T(t)}=-\lambda$$

So we get the following system of ordinary differential equations:

$\left\{\begin{matrix}
p(x) X''(x)+p'(x) X'(x)+ \lambda r(x) X(x)=0\\
T''(t)+\lambda T(t)=0
\end{matrix}\right.$

Can we find somehow the general form of the solution of this DE:

$p(x) X''(x)+p'(x) X'(x)+ \lambda r(x) X(x)=0$ ?

First, recognize that $p(x)X''(x)+ p'(x)X'(x)= (p(x)X'(x))'$. Whether we can find the general form of the solution depends upon the functions p(x) and r(x). This might be "Bessel's equation" or "Laguerre's equation", etc.

Also by solving the system, we will have found one solution of the pde, we will not know if the solution will be unique, right?

On the contrary, we know that the solution is not unique because a second order differential equation typically needs two additional conditions to determined the two undetermined constants you get when integrating.

Would the same happen if we would also have two boundary consitions as for $x$ and an initial condition as for $t$?

Since the original pde involved second derivatives with respect to both x and t, you will need two additional conditions for each. If, by "initial conditions", you mean **both** position and speed at t= 0, yes, that is sufficient.

or the second pde:

Suppose that $u(x,y)= X(x) Y(y)$. Then $u_{xx}+(x+y) u_{yy}=0 \Rightarrow X''(x) Y(y)+ (x+y) X(x) Y''(y)=0$.

So this cannot be solved with the method of separation of variables. And so we deduce that the solution isn't of the form $X(x) Y(y)$, right? How can we justify it more formally that this method cannot be used?

I would not jump to the conclusion that it cannot! What if you converted the problem to the independent variables x and z= x+ y?

For the last pde, I found the following system: $\left\{\begin{matrix}
X''+(x-\lambda)X=0\\
Y''+ \lambda Y=0
\end{matrix}\right.$

We don't know a genaral methodology to solve the first DE of the latter system, do we?

Because the first equation does not have constant coefficients? When you took "Ordinary Differential Equations", did you not learn "series solutions" methods for such equations?

 

[evinda]

Re: Method of separation of variables

First, recognize that $p(x)X''(x)+ p'(x)X'(x)= (p(x)X'(x))'$. Whether we can find the general form of the solution depends upon the functions p(x) and r(x). This might be "Bessel's equation" or "Laguerre's equation", etc.

In order to get the "Laguerre's equation" shouldn't we have $p(x)=x$ and $p'(x)=1-x$ ? But this can't be... (Worried) Also to get the "Bessel's equation" shouldn't the coefficient of $X'$ of the differential equation that we have be $\frac{p'(x)}{2}$ ?

On the contrary, we know that the solution is not unique because a second order differential equation typically needs two additional conditions to determined the two undetermined constants you get when integrating.

Ok.

Since the original pde involved second derivatives with respect to both x and t, you will need two additional conditions for each. If, by "initial conditions", you mean **both** position and speed at t= 0, yes, that is sufficient.

But then do we know that there is no other solution? I mean a solution that is not of the form $X(x) T(t)$ ?

I would not jump to the conclusion that it cannot! What if you converted the problem to the independent variables x and z= x+ y?

Let $z=x+y$.

We have that $$\frac{dX}{dx}=\frac{dX}{dz}\cdot \frac{dz}{dx}=\frac{dX}{dz}$$ and $$\frac{d^2X}{dx^2}=\frac{d}{dx}\left (\frac{dX}{dz}\right )=\frac{d}{dz}\frac{dX}{dz}\cdot \frac{dz}{dx}=\frac{d^2X}{dz^2}$$

Then we have $$\frac{d^2X}{dx^2}\cdot Y+(x+y)\cdot X\cdot \frac{d^2Y}{dy^2}=0 \\ \Rightarrow \frac{d^2X}{dz^2}\cdot Y+z\cdot X\cdot \frac{d^2Y}{dy^2}=0 \\ \Rightarrow \frac{d^2X}{dz^2}\cdot Y=-z\cdot X\cdot \frac{d^2Y}{dy^2} \\ \Rightarrow \frac{1}{z\cdot X}\cdot \frac{d^2X}{dz^2}=- \frac{1}{Y}\cdot \frac{d^2Y}{dy^2}$$

But won't $X$ be a variable of both $y$ and $z$ since $x=z-y$?

Because the first equation does not have constant coefficients? When you took "Ordinary Differential Equations", did you not learn "series solutions" methods for such equations?

So if we suppose that $X(x)=\sum_{n=0}^{\infty} a_n x^n$ then we will get that $\sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n+ \sum_{n=0}^{\infty} a_n x^{n+1}- \lambda \sum_{n=0}^{\infty} a_n x^n=0$, right?

And then we get information about the coefficients, right?

 

[HallsofIvy]

Re: Method of separation of variables

But then do we know that there is no other solution? I mean a solution that is not of the form $X(x) T(t)$ ?

There certainly are such solutions. They would be of the form $$\sum X(x)T(t)$$ where the sum is taken over all possible "$$\lambda$$". I'm surprised you did not learn that when you first learned about "separation of variables".