# Homework Help: Very difficult partial differential equation

1. Oct 12, 2007

### John1987

Last week, the professor gave a difficult PDE to solve as a bonus exercise, describing the motion of a conveyor belt. From experience, he knew that only 5% of the students (applied physics) is able to solve this problem. I got stuck and I really hope to get some help on this forum. This is the problem:

1. The problem statement, all variables and given/known data

Consider the conveyor belt equation:

$$u_{tt}+2*V*u_{tx}+(V^{2}-c^{2})*u_{xx}$$=0, 0<x<L and t>0. (eq. 1)

Furthermore, V,L and c are all constants and u= u(x,t). Boundary conditions: u(0,t)=u(L,t)=0

The question: construct nontrivial solutions of this problem in the form X(x)*T(t). He gave as a hint: "substitute X(x)*T(t) into the PDE and after dividing by X(x)*T(t), differentiate with respect to t or x.

2. Solution attempt

Well, this is how I started:

From the boundary conditions it follows that X(0)=X(L) = 0, since otherwise the solution would be trivial.

Substitution of u(x,t)= X(x)*T(t) into the PDE gives:

$$X(x)*T''(t)+2*V*X'(x)*T'(t)+(V^{2}-c^{2})*X''(x)*T(t)=0$$ (eq. 2)

Dividing by X(x)*T(t) yields:

$$\frac{T''(t)}{T(t)}+\frac{2*V*X'(x)*T'(t)}{X(x)*T(t)}+(V^{2}-c^{2})*\frac{X''(x)}{X(x)}=0$$ (eq. 3)

Now, I follow the professor's hint and differentiate this equation with respect to x. This gives:

$$2*V*\frac{T'(t)}{T(t)}*(X(x)*X''(x)-X'(x)*X'(x))+(V^{2}-c^{2})*(X(x)*X'''(x)-X''(x)*X'(x))=0$$ (eq. 4)
In this last step, I already multiplied both sides of the equation by $$X(x)^{2}$$ which appeared after applying the quotient rule for fraction differentiation. This last expression doesn't seem to help me so I also differentiate the original equation w.r.t. t. This gives:

$$T(t)*T'''(t)-T''(t)*T'(t)+2*V*\frac{X'(x)}{X(x)}*(T(t)*T''(t)-T'(t)*T'(t))=0$$ (eq. 5)
Here, I already multiplied out the factor $$T(t)^{2}$$ that came from applying the quotient differentiation rule.

This is the point where I get stuck: following the hints, I ended up with two nasty equations, neither one I can solve and it also seems very hard to use one of the two equations for solving the other.

Does anyone know how to proceed?

Thanks a lot,

John

Last edited: Oct 12, 2007
2. Oct 12, 2007

### dextercioby

Okay, use the second equation. Can you deduce that X'=C X ...?

3. Oct 12, 2007

### John1987

Well, I don't see how to to do that right now. Also I'm not completely sure which equation you mean. Is it the second equation, as I just numbered them?

4. Oct 13, 2007

### John1987

From equation 4, I can now deduce that $$\frac{T'(t)}{T(t)}$$ should be constant. This is because from eq.4, it follows that:

$$\frac{T'(t)}{T(t)}=\frac{1}{2*V}*\frac{(c^{2}-V^{2})*(X(x)*X'''(x)-X''(x)*X'(x))}{X*X''(x)-X'(x)*X'(x)}$$

The left hand side depends only on t and the right hand side only on x. By differentiation, it follows that indeed $$\frac{T'(t)}{T(t)}$$ is constant. Analogously, from equation 5 I know that $$\frac{X'(x)}{X(x)}$$ is also constant.

However, this implies that both X(x) and T(t) are exponential functions and this seems strange to me, especially given the boundary conditions X(0)=X(L)=0.

Can someone please say what I'm doing wrong here?

5. Oct 13, 2007

### Hurkyl

Staff Emeritus
Why do you think you can divide?

Let me simplify your work a bit to make things more clear -- if you differentiate equation (3) with respect to both x and t, you get

$$2 V \left( \frac{T'(t)}{T(t)} \right)' \left( \frac{X'(x)}{X(x)} \right)' = 0.$$

From this, the best we can conclude that one of V, $(T'(t)/T(t))'$, and $(X'(x)/X(x))'$ are zero.

You should never divide without first guaranteeing that the divisor is nonzero. (note that this can be accomplished by splitting into two cases: one where it is zero and one where it is not zero)

Last edited: Oct 13, 2007
6. Oct 13, 2007

### John1987

OK, this is quite helpful, thanks.

I know that V is nonzero from the information given in the question.

Also, $$\left( \frac{X'(x)}{X(x)} \right)'$$ cannot be zero since this implies that X(x) is given by an exponential function, which gives, after applying the boundary conditions, a trivial solution, which I'm not looking for.

So, $$\left( \frac{T'(t)}{T(t)} \right)'$$ must be zero, which gives me:

$$T(t)=(constant)*e^{\lambda*t}$$, where $$\lambda$$ is an arbitrary constant.

Last edited: Oct 13, 2007
7. Oct 13, 2007

### John1987

Does anyone know how to determine the constant $$\lambda$$ in the above equation? Usually, this can be done immediately after applying separation of variables, like in the Laplace equation. But now this doesn't seem to work.

8. Oct 13, 2007

### Hurkyl

Staff Emeritus
Well, how far have you gotten?

9. Oct 14, 2007

### John1987

Well, I know that, as derived earlier:

$$T(t)=(constant)*e^{\lambda*t}$$, where $$\lambda$$ is an arbitrary constant.

For T' and T'' I get:

$$T'(t)=\lambda*constant*e^{\lambda*t}$$ and $$T''(t)=\lambda^{2}*constant*e^{\lambda*t}$$

By substituting in eq.2, I get, after dividing the total equation by $$(constant)*e^{\lambda*t}$$:

$$X(x)*\lambda^{2}+2*V*X'(x)*\lambda+(V^{2}-c^{2})*X''(x)=0$$

This is an ordinary differential equation in X(x) and this is quite easy to solve. First I rewrite the equation for X(x) to get a characteristic equation. This gives:

$$r^{2}+\frac{2*\lambda*V}{V^{2}-c^{2}}*r+\frac{\lambda^{2}}{V^{2}-c^{2}}=0$$ (here I assume V is not equal to c)

Using the quadratic formula, it follows that there are two real roots:

r = $$\frac{-\lambda*V}{V^{2}-c^{2}}+\frac{\lambda*c}{V^{2}-c^{2}}$$

and r = $$\frac{-\lambda*V}{V^{2}-c^{2}}-\frac{\lambda*c}{V^{2}-c^{2}}$$

For the solution for X(x), this implies:

$$X(x)=c1*e^{(\frac{-\lambda*V}{V^{2}-c^{2}}+\frac{\lambda*c}{V^{2}-c^{2}})*x}+c2*e^{(\frac{-\lambda*V}{V^{2}-c^{2}}-\frac{\lambda*c}{V^{2}-c^{2}})*x}$$

Using the boundary conditions, X(0)=X(L)=0, I get that both c1 and c2 should be zero, which gives a trivial solution! Furthermore, I expected a sine or cosine in my solution for X(x) or T(t), but for some reason, both gave exponential terms ans no sines or cosines.

Please, can anyone help me with this exercise because it it quite annoying to get stuck every time you think to have gotten a correct answer.

Thanks,

John

Last edited: Oct 14, 2007
10. Oct 14, 2007

### Hurkyl

Staff Emeritus
There is one actual mistake: I think your computation for r is wrong. It looks like you squared something that shouldn't've been squared.

Oh, and why do you think your solution for r is real?

Nor do I see why c1 and c2 must both be zero, even if your r's are real.

Last edited: Oct 14, 2007
11. Oct 14, 2007

### John1987

My computation for r:

$$r = -\frac{\lambda*V}{V^{2}-c^{2}}\pm\frac{1}{2}*\sqrt{\frac{4*\lambda^{2}*V^{2}}{(V^{2}-c^{2})^{2}}-\frac{4*\lambda^{2}*(V^{2}-c^{2})}{(V^{2}-c^{2})^{2}}}$$

Simplifying gives me:

$$r = -\frac{\lambda*V}{V^{2}-c^{2}}\pm\frac{1}{2}*\sqrt{\frac{4*\lambda^{2}*c^{2}}{(V^{2}-c^{2})^{2}}$$

From this last equation, I already draw the conclusion that r is real:within the square root sign, negative numbers cannot appear.

Simplifying the last equation for r further, gives:

$$r=-\frac{\lambda*V}{V^{2}-c^{2}}\pm\frac{\lambda*c}{V^{2}-c^{2}}$$

From my earlier courses in ordinary DE's, I know that the general solution for X(x) can now be written as:

$$X(x)=c1*e^{(\frac{-\lambda*V}{V^{2}-c^{2}}+\frac{\lambda*c}{V^{2}-c^{2}})*x}+c2*e^{(\frac{-\lambda*V}{V^{2}-c^{2}}-\frac{\lambda*c}{V^{2}-c^{2}})*x}$$

Where the exponents are thus real.

BC X(0)=0 gives:

c1+c2=0

BC X(L)=0 gives:

$$0=c1*e^{(\frac{-\lambda*V}{V^{2}-c^{2}}+\frac{\lambda*c}{V^{2}-c^{2}})*L}+c2*e^{(\frac{-\lambda*V}{V^{2}-c^{2}}-\frac{\lambda*c}{V^{2}-c^{2}})*L}$$

Dividing the equation by

$$e^{\frac{-\lambda*V*L}{V^{2}-c^{2}}$$ and substituting c1=-c2 gives:

$$c2*e^{\frac{\lambda*c*L}{V^{2}-c^{2}}}-c2*e^{\frac{-\lambda*c*L}{V^{2}-c^{2}}=0$$

Finally, I rewrite this as:

$$2*c2*sinh(\frac{\lambda*c*L}{V^{2}-c^{2}})=0$$

This can only be true if c2=0 (since the argument of the sinh is nonzero) and that gives me the trivial solution, so I'm stuck again.

Last edited: Oct 14, 2007
12. Oct 14, 2007

### Hurkyl

Staff Emeritus
Aha! You do have the same thing I got -- the way I did it made it obvious that you can cancel out one of the factors of the bottom (which factor depends on which sign), and that's why your solution looks different than mine.

But I still stand by this objection. I was hoping not to spoil the answer... but $\lambda$ doesn't have to be real, does it?

Last edited: Oct 14, 2007
13. Oct 14, 2007

### Hurkyl

Staff Emeritus
And for similar reasons, I don't buy this.

14. Oct 14, 2007

### Hurkyl

Staff Emeritus
It strikes me that, when considering the case that (T' / T) is a constant, you neglected the case that T' / T is zero.

Or, which I think is equivalent, you neglected the case that your quadratic equation for r has a double root.

To get complete answers to differential equations, you really need to be a stickler for details. Each assumption you make will often prevent you from finding some class of "exceptional" solutions. (And the problem is especially pronounced if all solutions are exceptional!)

P.S. is u required to be real-valued? I've been assuming that wasn't a requirement.

Last edited: Oct 14, 2007
15. Oct 14, 2007

### Hurkyl

Staff Emeritus
I've nearly convinced myself that, subject to the constraints:

. u is nonzero
. u is real
. u has continuous second derivatives
. u(x, t) = X(x) T(t)
. X is never zero inside the strip
. T is never zero inside the strip
. L is nonzero
. c is nonzero
. V is nonzero
. V^2 - c^2 is nonzero

this pde doesn't have a solution.

I don't recall the properties of singular ode's to deal with the case that X or T may be zero inside the strip. I wonder if we can work around that problem by using your original equation (4), and not equation (5)? (which ought to be equivalent to the equation I posted) (Oh wait, I misread what you said there; I thought you said that you differentiated (4) w.r.t. t. No wonder it didn't look quite right. Bleh. )

Last edited: Oct 14, 2007
16. Oct 14, 2007

### Hurkyl

Staff Emeritus
Well, I suppose the best way to avoid singularities is not to divide, and start with

$$2 V (X(x) X''(x) - X'(x)^2 ) (T(t) T''(t) - T'(t)^2) = 0$$

$$2 V \left( \frac{X'(t)}{X(t)} \right)' \left( \frac{T'(t)}{T(t)} \right)' = 0.$$

I'm upset I didn't notice this sooner.

17. Oct 15, 2007

### John1987

Look, I'm trying to find non trivial solutions here, which implies that neither X(x) nor T(t) may be zero, since I assume: u(x.t)=X(x)*T(t). Given this constraint, it is allowed to divide by them.

So, in my opinion the equation $$2 V \left( \frac{X'(t)}{X(t)} \right)' \left( \frac{T'(t)}{T(t)} \right)' = 0.$$ would be a good one to work with.

18. Oct 15, 2007

### Hurkyl

Staff Emeritus
I think I was misunderstood. Given the hypothesis that u(t, x) = T(t) X(x), I can only prove prove that
$$2 V \left( \frac{X'(t)}{X(t)} \right)' \left( \frac{T'(t)}{T(t)} \right)' = 0$$
if I make the assumption that X and T are everywhere nonzero -- there cannot even be a single point in the strip where either of them is zero.

The problem is that I do not recall the properties of singular differential equations -- so in order to make this proof, I have to assume that dividing by X and T does not create any singularities; i.e. I have to assume that they cannot be zero at any single point.

However, I can prove
$$2 V (X(x) X''(x) - X'(x)^2 ) (T(t) T''(t) - T'(t)^2) = 0$$
without that assumption; any solution to $T(t) T''(t) - T'(t)^2 = 0$ should lead to a solution to the original differential equation. Clearly an exponential is a solution to this equation, but I cannot prove it is the only solution.

Last edited: Oct 15, 2007
19. Oct 15, 2007

### John1987

Well, assuming that this equation is correct:

$$2 V \left( \frac{X'(t)}{X(t)} \right)' \left( \frac{T'(t)}{T(t)} \right)' = 0.$$

We know that either $$\left( \frac{X'(t)}{X(t)} \right)'$$ or $$\left( \frac{T'(t)}{T(t)} \right)'$$ should be zero. I think you'll agree with me on that.

Assuming that $$\lambda$$ is real, I end up with a trivial solution, which is not what I'm looking for.

However, when I assume that $$\lambda$$ is complex-valued, then how do I solve the equation $$\left( \frac{T'(t)}{T(t)} \right)=\lambda$$????

20. Oct 15, 2007

### John1987

By the way: a hint was given together with the question that stated that it is allowed to divide by X(x)*T(t).