Very difficult partial differential equation

In summary, last week, the professor gave a difficult PDE problem involving the motion of a conveyor belt. He stated that only 5% of the students in applied physics are able to solve it. A student got stuck and sought help on a forum. The problem involves constructing nontrivial solutions in the form of X(x)*T(t) for the equation u_{tt}+2*V*u_{tx}+(V^{2}-c^{2})*u_{xx}=0, with boundary conditions u(0,t)=u(L,t)=0. The student attempted to solve the problem using the professor's hint, but got stuck and asked for assistance. The conversation ends with the student still struggling to solve the problem.
  • #1
John1987
12
0
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:

Homework Statement



Consider the conveyor belt equation:

[tex]u_{tt}+2*V*u_{tx}+(V^{2}-c^{2})*u_{xx}[/tex]=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:

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

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

[tex]\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[/tex] (eq. 3)

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

[tex]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[/tex] (eq. 4)
In this last step, I already multiplied both sides of the equation by [tex]X(x)^{2}[/tex] 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:

[tex]T(t)*T'''(t)-T''(t)*T'(t)+2*V*\frac{X'(x)}{X(x)}*(T(t)*T''(t)-T'(t)*T'(t))=0[/tex] (eq. 5)
Here, I already multiplied out the factor [tex]T(t)^{2}[/tex] 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
 
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  • #2
Okay, use the second equation. Can you deduce that X'=C X ...?
 
  • #3
dextercioby said:
Okay, use the second equation. Can you deduce that X'=C X ...?

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

[tex]\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)}[/tex]

The left hand side depends only on t and the right hand side only on x. By differentiation, it follows that indeed [tex]\frac{T'(t)}{T(t)}[/tex] is constant. Analogously, from equation 5 I know that [tex]\frac{X'(x)}{X(x)}[/tex] 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
John1987 said:
This is because from eq.4, it follows that:

[tex]\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)}[/tex]
...
Can someone please say what I'm doing wrong here?
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

[tex]
2 V \left( \frac{T'(t)}{T(t)} \right)' \left( \frac{X'(x)}{X(x)} \right)' = 0.
[/tex]

From this, the best we can conclude that one of V, [itex](T'(t)/T(t))'[/itex], and [itex](X'(x)/X(x))'[/itex] 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)
 
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  • #6
OK, this is quite helpful, thanks.

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

Also, [tex]\left( \frac{X'(x)}{X(x)} \right)' [/tex] 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, [tex]\left( \frac{T'(t)}{T(t)} \right)' [/tex] must be zero, which gives me:

[tex]T(t)=(constant)*e^{\lambda*t}[/tex], where [tex]\lambda[/tex] is an arbitrary constant.
 
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  • #7
Does anyone know how to determine the constant [tex]\lambda[/tex] 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. :confused:
 
  • #8
John1987 said:
Does anyone know how to determine the constant [tex]\lambda[/tex] 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. :confused:
Well, how far have you gotten?
 
  • #9
Hurkyl said:
Well, how far have you gotten?

Well, I know that, as derived earlier:

[tex]T(t)=(constant)*e^{\lambda*t}[/tex], where [tex]\lambda[/tex] is an arbitrary constant.

For T' and T'' I get:

[tex]T'(t)=\lambda*constant*e^{\lambda*t}[/tex] and [tex]T''(t)=\lambda^{2}*constant*e^{\lambda*t}[/tex]

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

[tex]X(x)*\lambda^{2}+2*V*X'(x)*\lambda+(V^{2}-c^{2})*X''(x)=0[/tex]

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:

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

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

r = [tex]\frac{-\lambda*V}{V^{2}-c^{2}}+\frac{\lambda*c}{V^{2}-c^{2}}[/tex]

and r = [tex]\frac{-\lambda*V}{V^{2}-c^{2}}-\frac{\lambda*c}{V^{2}-c^{2}}[/tex]

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

[tex]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}[/tex]

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
 
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  • #10
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.
 
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  • #11
Hurkyl said:
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.

My computation for r:

[tex]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}}}[/tex]

Simplifying gives me:

[tex]r = -\frac{\lambda*V}{V^{2}-c^{2}}\pm\frac{1}{2}*\sqrt{\frac{4*\lambda^{2}*c^{2}}{(V^{2}-c^{2})^{2}}[/tex]

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:

[tex]r=-\frac{\lambda*V}{V^{2}-c^{2}}\pm\frac{\lambda*c}{V^{2}-c^{2}}[/tex]

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

[tex]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}[/tex]

Where the exponents are thus real.

BC X(0)=0 gives:

c1+c2=0

BC X(L)=0 gives:

[tex]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}[/tex]

Dividing the equation by

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

[tex]c2*e^{\frac{\lambda*c*L}{V^{2}-c^{2}}}-c2*e^{\frac{-\lambda*c*L}{V^{2}-c^{2}}=0[/tex]

Finally, I rewrite this as:

[tex]2*c2*sinh(\frac{\lambda*c*L}{V^{2}-c^{2}})=0[/tex]

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.
 
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  • #12
Hurkyl said:
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.
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.

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



Oh, and why do you think your solution for r is real?
But I still stand by this objection. :smile: I was hoping not to spoil the answer... but [itex]\lambda[/itex] doesn't have to be real, does it?
 
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  • #13
John1987 said:
Also, [tex]\left( \frac{X'(x)}{X(x)} \right)' [/tex] 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.
And for similar reasons, I don't buy this.
 
  • #14
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.
 
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  • #15
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. :frown: 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. :frown:)
 
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  • #16
Well, I suppose the best way to avoid singularities is not to divide, and start with

[tex]2 V (X(x) X''(x) - X'(x)^2 ) (T(t) T''(t) - T'(t)^2) = 0[/tex]

instead of

[tex]2 V \left( \frac{X'(t)}{X(t)} \right)' \left( \frac{T'(t)}{T(t)} \right)' = 0.[/tex]

I'm upset I didn't notice this sooner.
 
  • #17
Hurkyl said:
Well, I suppose the best way to avoid singularities is not to divide, and start with

[tex]2 V (X(x) X''(x) - X'(x)^2 ) (T(t) T''(t) - T'(t)^2) = 0[/tex]

instead of

[tex]2 V \left( \frac{X'(t)}{X(t)} \right)' \left( \frac{T'(t)}{T(t)} \right)' = 0.[/tex]

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

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 [tex]2 V \left( \frac{X'(t)}{X(t)} \right)' \left( \frac{T'(t)}{T(t)} \right)' = 0.[/tex] would be a good one to work with.
 
  • #18
I think I was misunderstood. Given the hypothesis that u(t, x) = T(t) X(x), I can only prove prove that
[tex]2 V \left( \frac{X'(t)}{X(t)} \right)' \left( \frac{T'(t)}{T(t)} \right)' = 0[/tex]
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
[tex]2 V (X(x) X''(x) - X'(x)^2 ) (T(t) T''(t) - T'(t)^2) = 0[/tex]
without that assumption; any solution to [itex]T(t) T''(t) - T'(t)^2 = 0[/itex] 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. :frown:
 
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  • #19
Hurkyl said:
But I still stand by this objection. :smile: I was hoping not to spoil the answer... but [itex]\lambda[/itex] doesn't have to be real, does it?


Well, assuming that this equation is correct:

[tex]2 V \left( \frac{X'(t)}{X(t)} \right)' \left( \frac{T'(t)}{T(t)} \right)' = 0.[/tex]

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

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

However, when I assume that [tex]\lambda[/tex] is complex-valued, then how do I solve the equation [tex]\left( \frac{T'(t)}{T(t)} \right)=\lambda[/tex]?
 
  • #20
Hurkyl said:
I think I was misunderstood. Given the hypothesis that u(t, x) = T(t) X(x), I can only prove prove that
[tex]2 V \left( \frac{X'(t)}{X(t)} \right)' \left( \frac{T'(t)}{T(t)} \right)' = 0[/tex]
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
[tex]2 V (X(x) X''(x) - X'(x)^2 ) (T(t) T''(t) - T'(t)^2) = 0[/tex]
without that assumption. Clearly the exponential is a solution to [itex]T(t) T''(t) = T'(t)^2[/itex], but I cannot yet prove that the exponential is the only solution.

By the way: a hint was given together with the question that stated that it is allowed to divide by X(x)*T(t).
 
  • #21
John1987 said:
However, when I assume that [tex]\lambda[/tex] is complex-valued, then how do I solve the equation [tex]\left( \frac{T'(t)}{T(t)} \right)=\lambda[/tex]?
It's exactly the same: [itex]T(t) = A e^{\lambda t}[/itex].
 
  • #22
Hurkyl said:
It's exactly the same: [itex]T(t) = A e^{\lambda t}[/itex].

That I know, but in that case, I get a complex-valued function u(x,t), right? I don't think Euler's equation can now be used to write this as sines and cosines, because I have only one imaginary "root" to my characteristic equation [tex]r-\lambda=0[/tex].
 
  • #23
did u manage to find a solution to this problem..?
 
  • #24
Could anyone explain a little bit of the physical meaning of this PDE?

It should be relevant to vibration equation...

Thank you in advance.
 
  • #25
@Hurkyl: did you find the solution (2 years later?).
 
  • #26
I think I solved it. You don't need to solve that very difficult DE for X(x). Just substitue the expression you'll get for the separation constant in the original DE and solve the newly formed ODE.
 
  • #27
@dirk_mec1

How and what expression do you get for the separation constant. I know that usually this is done by the boundary conditions, but I don't see how that works here, because application of the boundary conditions imply a trivial solution .
 
  • #28
Mrx said:
@dirk_mec1
How and what expression do you get for the separation constant.
Just follow the hint. Seperate variables, differentiate w.r.t x or t and then substitute the expression for [tex] \lambda [/tex] in the original PDE.

I know that usually this is done by the boundary conditions, but I don't see how that works here, because application of the boundary conditions imply a trivial solution .

Just show me what you've got and I'll help you.
 
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  • #29
After seperating variables and deviding by XT:

[tex]\frac{T''}{T} + 2V\frac{X'T'}{XT} + (V^2-c^2) \frac{X''}{X}=0[/tex]

differentiation wrt to t:

[tex]\frac{d}{dt}\left(\frac{T''}{T}\right) = -2V\frac{d}{dt}\left(\frac{X'T'}{XT}\right)[/tex] (1)

differentiating wrt to x:

[tex](V^2 -c^2)\frac{d}{dx}\left(\frac{X''}{X}\right) = -2V\frac{d}{dx}\left(\frac{X'T'}{XT}\right)[/tex] (2)

for example 1st equation:

[tex]\frac{\frac{d}{dt}\left(\frac{T''}{T}\right) }{\frac{d}{dt}\left(\frac{T'}{T}\right)}= -2V\frac{X'}{X} = -\lambda[/tex]

where [tex]\frac{d}{dt}\left(\frac{T'}{T}\right)\neq 0[/tex]

However solving the time dependent part leads to [tex]\frac{d}{dt}\left(\frac{T'}{T}\right) = \frac{d}{dt}\left(- \lambda\right) = 0[/tex] so deviding by [tex]\frac{d}{dt}\left(\frac{T'}{T}\right)[/tex] is not justified.

Then I followed the directions given in this thread to prove that [tex]0=\frac{d}{dx}\left(\frac{X'}{X}\right)\frac{d}{dt}\left(\frac{T'}{T}\right)[/tex] Which is easy to see by differentiating eq. 1 w.r.t. x or eq. 2 w.r.t. t.

So suppose [tex]\frac{d}{dt}\left(\frac{T'}{T}\right)=0[/tex], then [tex]T(t)=\exp{-\lambda t}[/tex]. Substitution in the separated PDE gives a ODE with solution

[tex]X(x)=A\exp{\lambda x /(V+c)}+B\exp{\lambda x/(V-c)}[/tex]

However applying the boundary conditions [tex]X(0)=X(L)=0[/tex] leads to the trivial solution [tex]X(x)=0[/tex]

So this leaves us [tex]\frac{d}{dx}\left(\frac{X'}{X}\right)=0[/tex]

is this correct so far?

_________________________
EDIT: I realize I was a not very careful in my last statement that it leads to [tex]X=0[/tex]. If [tex]\lambda[/tex] is purely imaginary then

[tex]X(x)=A\exp\left(ax\right)\sin(bx)[/tex]

with a and b some constants. This ofcourse does satisfy the boundary conditions. However is it okay to assume that [tex]\lambda[/tex] is purely imaginary?
 
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  • #30
After separating variables and dividing by XT:

[tex]
\frac{T''}{T} + 2V\frac{X'T'}{XT} + (V^2-c^2) \frac{X''}{X}=0
[/tex]


differentiation w.r.t. x

[tex]
(V^2 -c^2)\frac{d}{dx}\left(\frac{X''}{X}\right) = -2V\frac{d}{dx}\left(\frac{X'T'}{XT}\right) \rightarrow \frac{(V^2 -c^2)\frac{d}{dx}\left(\frac{X''}{X}\right)}{ (-2V) \frac{d}{dx} \left( \frac{X'}{X} \right) } = \frac{T'}{T}= \lambda
[/tex]



So:

[tex] T'' = \lambda T' = \lambda T^2 [/tex]



Substituting into the original PDE gives:

\lambda ^2 X + 2V \lambda X' + (V^2 -c^2) X" = 0

(Latex is doing weird things here)
 
  • #31
Thanks for your reply. The equation you got for X(x) is the same I tried to solve. Let's do it again:

[tex]\lambda^2+2VX'\lambda + (V^2-c^2)X''=0[/tex]

has solution

[tex]X(x)=\exp\left(r_{\pm} x\right)[/tex]

With [tex]r=\frac{V\lambda \pm c\sqrt{\lambda^2}}{V^2-c^2}[/tex]

Now you can distinguish three cases:
  • [tex]\lambda[/tex] purely real
  • [tex]\lambda[/tex] purely imaginary
  • [tex]\lambda[/tex] = a + bi

[tex]\lambda[/tex] purely real
Then
[tex]r=\frac{-V\lambda \pm c\lambda}{V^2-c^2}=\frac{-V\lambda \pm c\lambda}{(V+c)(V-c)}=\frac{-\lambda}{V \pm c}[/tex]
So
[tex]X(x)=A\exp\left(\frac{-\lambda}{V+c}x\right)+B\exp\left(\frac{-\lambda}{V-c}x\right)[/tex]
With boundary conditions: [tex]X(0)=X(L)=0[/tex] it follows that A + B = 0 and
[tex]X(L)=A\exp\left(\frac{-\lambda}{V+c}L\right)-A\exp\left(\frac{-\lambda}{V-c}L\right)=0\Leftrightarrow A=0[/tex]

It doesn't matter which values lambda takes ([tex]\lambda>0,\lambda<0, \lambda=0[/tex]). [tex]\lambda[/tex] cannot be purely real.

[tex]\lambda[/tex] purely imaginary
Say [tex] \lambda = b i [/tex] Then [tex] r=\frac{-b V i \pm c b i}{V^2-c^2}[/tex].
And
[tex]X(x)=A\exp\left(\frac{-b V i + c b i}{V^2-c^2}x \right)+B\exp\left(\frac{-b V i - c b i}{V^2-c^2}x \right)[/tex]
[tex]X(0)=0[/tex] gives: B=-A, so
[tex]X(x)=A\exp\left(\frac{-bVi}{V^2-c^2}x \right) \left(\exp\left(\frac{c b i}{V^2-c^2}x \right)-\exp\left(\frac{- c b i}{V^2-c^2}x \right)\right)[/tex]
[tex]X(x)=2iA\exp\left(\frac{-bVi}{V^2-c^2}x \right) \left(\sin\left( \frac{bcx}{V^2-c^2}\right))\right)[/tex]
[tex]X(L)=0[/tex] gives [tex]b=\frac{n\pi \left( V^2 -c^2 \right)}{cL}[/tex]
so [tex]\lambda = \frac{n\pi \left( V^2 -c^2 \right)}{cL} i [/tex]

In the end then:

[tex]u(x,t)= \sum_{n=1}^{\infty} a_n \sin\left(\frac{n \pi x}{L} \right)\exp\left(\frac{n\pi V x}{cL}i \right) \exp\left(\frac{n\pi (V^2-c^2) t}{cL}i \right)[/tex]

Did you get this as a solution?

I haven't tried the third case yet.
 
  • #32
Mrx said:
Thanks for your reply. The equation you got for X(x) is the same I
[tex]u(x,t)= \sum_{n=1}^{\infty} a_n \sin\left(\frac{n \pi x}{L} \right)\exp\left(\frac{n\pi V x}{cL}i \right) \exp\left(\frac{n\pi (V^2-c^2) t}{cL}i \right)[/tex]

Did you get this as a solution?

Yes, that's the solution I got! It's tough but we've made it! Note that you can write this in sine and cosine and you can determine the (in this case 2) coefficients via the initial conditions.
 
  • #33
Okay great! thank you for your help.

Do you know what this equation ectually represents? I can't find anything about the physical meaning, or a derivation of this equation. I tried searching on conveyor belt equation but got nothing.
 
  • #34
spinoza1989 said:
Could anyone explain a little bit of the physical meaning of this PDE?

If V=0 than you have the wave equation. So c is the speed of the wave (in the belt) and u is the displacement in the y-direction. So what does it means that V=0, I think it means the belt is not moving. What is very strange to me because V is a constant and not time depending.

In the solution given by
Mrx said:
[tex]u(x,t)= \sum_{n=1}^{\infty} a_n \sin\left(\frac{n \pi x}{L} \right)\exp\left(\frac{n\pi V x}{cL}i \right) \exp\left(\frac{n\pi (V^2-c^2) t}{cL}i \right)[/tex]

you can bring the exp-parts together and you get a x-v't where v' is a the speed of the wave relative to the frame. (c is speed relative to the belt and V the speed of the belt relative to the frame).

[tex]2 V u_x_t + V^2 u_x_x[/tex] is the "source therm" I think, first of all since you have just a wave equations and the other therms are source therms. Second reason is that the units of [tex]u_x_t [/tex] are Hertz and mabye it has to do with the rotations of one of the wheels.

The 2V has something to do with the speed between upper belt and lower (returning) belt.

This is all what I could thought about.
Maybe somebody can make a 3dplot in maple?

Greetz Bendavid2
 

1. What is a partial differential equation?

A partial differential equation (PDE) is a mathematical equation that involves multiple independent variables and their partial derivatives. It is used to describe relationships between different quantities in a system, such as the rate of change of temperature over time and space.

2. What makes a partial differential equation difficult?

A partial differential equation can be difficult due to its complexity and the fact that it involves multiple variables and their derivatives. Additionally, the behavior of solutions to PDEs can be unpredictable and may require advanced mathematical techniques to solve.

3. What are some real-world applications of very difficult partial differential equations?

PDEs are used in many fields of science and engineering, including physics, chemistry, biology, and economics. Some specific applications include modeling fluid flow, heat transfer, and electromagnetic fields.

4. How are very difficult partial differential equations solved?

Solving a PDE can be a challenging task and often requires advanced mathematical techniques, such as separation of variables, Fourier series, or numerical methods. In some cases, exact solutions may not be possible, and approximations or numerical simulations are used instead.

5. What skills are needed to work with very difficult partial differential equations?

To work with very difficult partial differential equations, one needs a strong background in mathematics, particularly in calculus, linear algebra, and differential equations. Knowledge of advanced techniques such as Fourier analysis and numerical methods is also helpful. Additionally, problem-solving skills and a strong understanding of the physical or mathematical system being studied are essential.

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