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) k2∂2U∂t2=∂2U∂x2
(2) U(0,t)=0
(3) U(L,t)=0
(4) U(x,0)=0
(5) ∂U∂t(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′′=X′′T
Dividing each side by
XT:
k2T′′T=X′′X
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.
k2T′′T=X′′X=constant={0−λ2λ2
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:
X′′X=0⇒X′′=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:
k2T′′T=0⇒T′′=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)=0⇒B(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)=0⇒L(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:
X′′X=−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:
k2T′′T=−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)=0⇒sin(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,∀n∈N
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)=0⇒sin(nπxL)[Ccos(0)+Dsin(0)]=0
cos(0)=1 and
sin(0)=0, so,
U(x,0)=0⇒sin(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)
∀n∈N, 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
∂U∂t:
∂U∂t=∞∑n=1Dnsin(nπxL)(nπLk)cos(nπLkt)
Looking at the actual constraint, (5):
∂U∂t(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)=2L∫L0f(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=2knπ∫L0f(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.