Partial Differential Equations - solving using separation of varaibles

Mar 2016
170
0
Australia
Hi

I have to solve an equation:

uxx- 4utt = 0

to find all possible product form solutions to the equation.


  • As we have done in examples I have tried u(x,t) = T(t) X(x) resulting in:

T''X = TX''


  • Dividing through by XT yields:

T''X/XT = TX''/XT which gives: T''/T = X''/X.


  • As I understand it this equality means that the only possible result is that both must be a constant. We then looked at the solution for c=0, c= +ve and c = -ve (when c is the constant - used lamda (?) but don't know how to type that.

The only examples we have done did not involve two second derivatives so for c = 0 we got a constant for one equation, say T' and a+bx for the other, say X''. Also we have not done an example when one of the terms has a co-efficient. I have looked through my text book and can't work out how to do this as there are no worked solutions so it is difficult to follow the steps.

Does that mean that in this case, if c = 0 then:

T'' = 0 therefore T' = Ax+b; and
X'' = 0 therefore X' = Ax+b.


  • What then is the case if c is positive, would I get a solution in the form: Ae^(cx) + Be^(cx) for both equations? If that is the case would the solution be both of the equations involving exponentials being multiplied together?
  • Does the same thing result if c was negative?

I would be grateful for any help with this.

Kind regards
Beetle
 

topsquark

Forum Staff
Jan 2006
11,568
3,453
Wellsville, NY
Hi

I have to solve an equation:

uxx- 4utt = 0

to find all possible product form solutions to the equation.


  • As we have done in examples I have tried u(x,t) = T(t) X(x) resulting in:

T''X = TX''


  • Dividing through by XT yields:

T''X/XT = TX''/XT which gives: T''/T = X''/X.


  • As I understand it this equality means that the only possible result is that both must be a constant. We then looked at the solution for c=0, c= +ve and c = -ve (when c is the constant - used lamda (?) but don't know how to type that.

The only examples we have done did not involve two second derivatives so for c = 0 we got a constant for one equation, say T' and a+bx for the other, say X''. Also we have not done an example when one of the terms has a co-efficient. I have looked through my text book and can't work out how to do this as there are no worked solutions so it is difficult to follow the steps.

Does that mean that in this case, if c = 0 then:

T'' = 0 therefore T' = Ax+b; and
X'' = 0 therefore X' = Ax+b.


  • What then is the case if c is positive, would I get a solution in the form: Ae^(cx) + Be^(cx) for both equations? If that is the case would the solution be both of the equations involving exponentials being multiplied together?
  • Does the same thing result if c was negative?

I would be grateful for any help with this.

Kind regards
Beetle
First, you forgot about the factor of 4.

Second, if T'' = 0, then the general solution would be T = at + b, not ax + b. Then we also have X = dx + e.

And yes, if c > 0 you have two exponential terms for each X(x) and T(t). etc. for c < 0.

-Dan
 
Mar 2016
170
0
Australia
Thank you,

Am I able to deal with the factor of 4 at the beginning and end up with x terms divided by 4 or do I deal with it at the solution stage - I am thinking it would effect the derivative so would need to carry it through to the solution stage.

Also, I really don't understand what the solution looks like - when we looked at the c = 0 case he left the solution as two separate expressions, one being the constant and one being a linear equation. When we looked at the exponential case he multiplied one by the other. So do I end up with 3 solutions - being a combined solution for each case or 6 solutions, being two solutions for each case.

Sorry for all the stupid questions I am just really struggling to follow what we went through in lectures and apply that to the question.

Kind regards
Beetle
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
The "4" comes from the original equation so you should include it "at the beginning". Taking \(\displaystyle u(x, t)= X(x)T(t)\) the equation \(\displaystyle u_{xx}- 4u_{tt}= 0\) becomes \(\displaystyle TX''- 4XT''= 0\), \(\displaystyle TX''= 4XT''\) and, dividing by XT as you do, \(\displaystyle \frac{X''}{X}= 4\frac{T''}{T}\). Yes, since the left side is a function of x only and the right side a function of t only, and x and t are independent variables, we could change t while not changing x. In that case, since x has not changed, the left side would stay constant which means the right side must stay constant even though t has changed: for all t \(\displaystyle 4\frac{T''}{T}\) is a constant. Changing x but not t similarly shows that \(\displaystyle \frac{X''}{X}\) is a constant. Calling that constant c we have \(\displaystyle \frac{X''}{X}= c\) so \(\displaystyle X''= cX\) and \(\displaystyle 4\frac{T''}{T}= 0\) so \(\displaystyle T''= \frac{1}{4}T\).

Now, yes, there are three distinctly different cases:
1) c= 0. So we have \(\displaystyle X''= 0\) and \(\displaystyle T''= 0\) so that X= ax+ b and T= ct+ d for constants a, b, c, and d.
2) c> 0. Since c is positive we can write \(\displaystyle c= \lambda^2\) for some real number \(\displaystyle \lambda\) so that \(\displaystyle c= \pm\lambda\). So we have \(\displaystyle X''= \lambda X\) and \(\displaystyle T''= \lambda T\). Each of those has characteristic equation \(\displaystyle r^2= \lambda^2\) which has roots \(\displaystyle \lambda\) and \(\displaystyle -\lambda\). The general solutions are \(\displaystyle X= ae^{\lambda t}+ be^{-\lambda t}\) and \(\displaystyle T= ce^{\lamba t}+ de^{\lambda t}\)
3) c< 0. Since c is negative, we can write \(\displaystyle c= -\lambda^2\) for some real number \(\displaystyle \lambda\) so that \(\displaystyle c= \pm \lambda i\) So we have \(\displaystyle X''= \lambda i X\) and \(\displaystyle T''= \lambda I T\). As before, we can write the solutions \(\displaystyle X= ae^{i\lambda x}+ be^{-i\lambda x}\) and \(\displaystyle T= ce^{i\lambda t}+ de^{-i\lambda t}\). Of course, exponentials with imaginary exponents can be written as sine and cosine, \(\displaystyle e^{iax}= cos(ax)+ isin(ax)\) so those can both be written as \(\displaystyle X= acos(\lambda x)+ bsin(\lambda x)\) and \(\displaystyle T= c cos(\lambda t)+ d sin(\lambda t)\).

But you have forgotten a very important part of the problem- The initial and boundary conditions! Those are what determine which case we are dealing with. For example, suppose the boundary conditions are that x(0)= 0 and x(L)= 0 for some non-zero number L. The only way \(\displaystyle X= ax+ b\) can be 0 at both x= 0 and x= L is if X is identically 0: X(0)= b= 0 and then X(L)= aL+ 0= 0 so a= 0. Similarly, the only way \(\displaystyle ae^{\lambda x}+ be^{-\lambda x}\) can satisfy X(0)= X(L)= 0 is if both a and b are 0: \(\displaystyle X(0)= ae^0+ be^0= a+ b= 0\) and \(\displaystyle X(L)= ae^{\lamba L}+ be^{-\lambda L}= 0. From a+ b= 0, b= -a so the second equation becomes \(\displaystyle ae^{\lambda L}- ae^{-\lambda L}= a(e^{\lambda L}- e^{-\lambda})= 0\). But \(\displaystyle e^{\lambda L}- e^{-\lambda L}\) can't be 0- one of the exponents is positive while the other negative so one o the exponential is greater than 1 while the other is less than 1. Again the we must have b= -a= 0.

But if \(\displaystyle X= acos(\lamba x)+ bsin(\lambda x)\) then \(\displaystyle X(0)= a= 0\) and \(\displaystyle X(L)= a cos(\lambda L)+ b sin(\lambda L)= b sin(\lambda L)= 0\) so either b= 0, giving the "trivial solution", X(x)= 0 for all x, or \(\displaystyle \lambda L\) is a multiple of \(\displaystyle \pi\). If \(\displaystyle \lambda L= n\pi\), for integer n, then \(\displaystyle \lambda= \frac{n \pi}{L}\) giving solution \(\displaystyle X(x)= b sin(\frac{n\pi}{L})\). The "b" is determined by the initial condition and the fact that n can be any integer gives rise to a sum over integers, the "Fourier series".

As for "when we looked at the c = 0 case he left the solution as two separate expressions, one being the constant and one being a linear equation. When we looked at the exponential case he multiplied one by the other", yes, you saw that in ordinary differential equations. If \(\displaystyle \frac{dy}{dx}= a\) so that \(\displaystyle dy= adx\) then, integrating, \(\displaystyle y= ax+ c\) where c is an arbitrary constant. If \(\displaystyle \frac{dy}{dx}= ay\) so that \(\displaystyle dy= ay dx\) we cannot immediately integrate since on the right we are integrating with respect to x but have y, an unknown function of x. Instead divide both sides by y: \(\displaystyle \frac{dy}{y}= a dx\) and then integrate to get \(\displaystyle ln(y)= ax+ c\). Solve for y by taking the exponential of both sides: \(\displaystyle y= e^{ax+ c}= e^ce^{ax}= Ce^{ax}\) where \(\displaystyle C= e^c\).\)
 
Last edited:
Mar 2016
170
0
Australia
Thanks, it is starting to make sense. We were not given any IC or BC so just need the general solutions. Kind regards beetle
 
Mar 2016
170
0
Australia
Hi

I am not sure if the attachment worked.

If it did is anyone able to provide any further guidance. The attached is what I have done with the help received so far which has been fantastic. However when I was working through it I found I still wasn't sure how to use the factor of 4. In relation to the third category of solutions we did not go through why the solution was in that form, rather just told to remember it as we weren't going into imaginary numbers.

I am trying to show I am doing the best I can do, not just asking for the answers as I need to learn how to do this.

Kind regards
Beetle
 

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