Wednesday, March 27, 2013

Simple Partial Differential Equations Example

I am really enjoying my current partial differential equations class, so I thought I'd share an example problem. Note that this is probably one of the simplest problems in partial DE.

Problem

(1) k22Ut2=2Ux2

(2) U(0,t)=0

(3) U(L,t)=0

(4) U(x,0)=0

(5) Ut(x,0)=f(x)


Solution


We are looking for a solution of the form:
U=XT

Where X is a function of x and T is a function of t.

Translating (1) to match the expected solution, we get:
k2XT=XT


Dividing each side by XT:
k2TT=XX


Since we have a function of only T on the left and only X on the right, we know that these are equal to a constant.
k2TT=XX=constant={0λ2λ2


The notation above says that the constant can either be 0, some negative number (as forced to be negative by λ2) or some positive number (forced by λ2).

Case constant=0:

Finding X:
XX=0X=0


Integrating both sides:
X=A

Where A is an arbitrary constant.

Integrating again:
X=Ax+B

Where B is an arbitrary constant.

Finding T:
k2TT=0T=0


Integrating both sides:
T=C

Where C is an arbitrary constant.

Integrating again:
T=Ct+D

Where D is an arbitrary constant.

Since U=XT:
U=(Ax+B)(Ct+D)

Therefore, this satisfies (1).

Looking at (2), we have:
U(0,t)=0(A(0)+B)(Ct+D)=0B(Ct+D)=0


This implies that B=0, so:
U=Ax(Ct+D)=x(Ct+D)

satisfies (1) and (2); note that the constant A was absorbed into the other constants (since they are, after all, just arbitrary constants).

Looking at (3), we have:
U(L,t)=0L(Ct+D)=0


To make this true, we can't change the value of L because it is a constraint, so the only option is to make C=D=0.

So,
U=0

This, however, is an uninteresting solution for U. So, we examine the next possible constant.

Case constant=λ2:

Finding X:
XX=λ2


Multiply both sides by X:
X=λ2X


Using methods from differential equations (DE), we know that we can solve this by subbing α2 in for X and 1 in for X:
α2=λ2


Taking the square root of both sides we get:
α=±λi


From DE we know that this fits the form α=b±ci, where the answer the the DE is:
X=ebx[Acos(cx)+Bsin(cx)]

Where A and B are arbitrary constants.

Therefore:
X=e0x[Acos(λx)+Bsin(λx)]=[Acos(λx)+Bsin(λx)]


Finding T:

k2TT=λ2


Multiply both sides by Tk2:
T=λ2k2T


Using methods from differential equations (DE), we know that we can solve this by subbing β2 in for T and 1 in for T:
β2=λ2k2


Taking the square root of both sides we get:
β=±λki


From DE we know that this also fits the form β=b±ci, where the answer the the DE is:
T=ebt[Ccos(ct)+Dsin(ct)]

Where C and D are arbitrary constants (not necessarily the same as A and B).

Therefore:
T=e0t[Ccos(λkt)+Dsin(λkt)]=[Ccos(λkt)+Dsin(λkt)]


Since U=XT:
U=[Acos(λx)+Bsin(λx)][Ccos(λkt)+Dsin(λkt)]

Therefore, this satisfies (1).

Looking at (2), we have:
U(0,t)=0[Acos(0)+Bsin(0)][Ccos(λkt)+Dsin(λkt)]=0


cos(0)=1 and sin(0)=0, so,
A[Ccos(λkt)+Dsin(λkt)]=0


This implies that A=0, so:
U=sin(λx)[Ccos(λkt)+Dsin(λkt)]
  
satisfies (1) and (2); note that the constant B was absorbed into the other constants (since they are, after all, just arbitrary constants).

Looking at (3), we have:
U(L,t)=0sin(λL)[Ccos(λkt)+Dsin(λkt)]=0


So, we need to pick a value for λ that causes the above expression to always be zero. We find that there are an infinite number of them (everywhere sin(nπ) where n is a natural number (better suited to match the solution than an integer). So, because of this, we find that:
λ=nπL,nN


So, subbing in for λ, we have:
U=sin(nπxL)[Ccos(nπLkt)+Dsin(nπLkt)]

Which satisfies (1), (2) and (3).

Looking at (4), we have:
U(x,0)=0sin(nπxL)[Ccos(0)+Dsin(0)]=0


cos(0)=1 and sin(0)=0, so,
U(x,0)=0sin(nπxL)(C)=0


This implies that C=0, so,
U=sin(nπxL)(D)sin(nπLkt)

Satisfies (1), (2), (3) and (4).

Since constraint (5) is non-zero, I will evaluate it lastly. For now, I claim that:
U=n=1Dnsin(nπxL)sin(nπLkt)

nN, also satisfies (1), (2), (3), and (4). This claim is justified because, as show above, constraint (3) is satisfied for all values of n. So, for whatever n I pick from the natural numbers (1,2,3..., N) the resulting U will also be an answer. So, summing together the infinite answers for U still meets the problem's criteria. D became Dn because D is not necessarily the same thing for any n value.

So, considering constraint (5), we need to first find Ut:
Ut=n=1Dnsin(nπxL)(nπLk)cos(nπLkt)


Looking at the actual constraint, (5):
Ut(x,0)=f(x)n=1Dnsin(nπxL)(nπLk)cos(0)=f(x)


Since, cos(0)=1:
f(x)=n=1Dnsin(nπxL)(nπLk)


By the Fourier sine series, we know that the coefficients are given by:
Dn(nπLk)=2LL0f(x)sin(nπxL)dx

We include (nπLk) because it does not match the Fourier series and needs to be divided out.

Solving for the actual coefficients:
Dn=2knπL0f(x)sin(nπxL)dx


Where our final U is, as above:
U=n=1Dnsin(nπxL)sin(nπLkt)


Now, we could go on to examine the constant=λ2 case. However, it ends up just being 0 (and is therefore redundant). You can use methods like those above to come to this conclusion for yourself.

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