Summary of Partial Differnetial Equations

Summary of Partial Differential Equations
Jonathan Pilgram
January 31, 2023


1 Eigenfunction techniques
The general idea is to define a linear set of eigenvalues with eigenfunctions that together solve the PDE.

1.1 Separation of variables
Take as an example the heat equation:
ut = kuxx
For:
0 < x < L; t > 0
Now we assume that:
u(x, t) = X(x)T (t)
Gives:
1 Tt Xxx
= = −λ
kT X
So we find that:
1 Tt
= −λ
kT
Tt = −λkT → T (t) = C0 e−λkt
And:
Xxx
= −λ → Xxx = −λX
X
Which is a second order ODE.

1.2 Sturm-Liouville problems
Define a linear differential operator:
   
1 d d
L=− p(x) + q(x)
σ(x) dx dx

Now one can write second order ordinary differential equations as:

Lu = −λu
 
d du
p(x) + q(x)u = −σ(x)λu
dx dx
A general solution is of the form:
X
cn (t)ϕn (x) = u(x, t)
n

with only positive eigenvalues (cn ).

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, 1.2.1 What if an ODE is not in Sturm-Liouville form
Then you can try a trick. Given an equation of the form:
d2 u(x) du(x)
+ α(x) = −λu(x)
dx2 dx
Which is not in Sturm-Liouville form, because the first order
∫ term is not combinable with the second
order term. Then we can multiply both sides by H(x) = e α(x) dx to get:
d2 u(x) du(x) d du(x)
2
H(x) + α(x)H(x) = −λH(x)u(x) → (H(x) ) = −λH(x)u(x)
dx dx dx dx
Which is Sturm-Liouville form (with σ(x) = H(x) and q(x) = 0)

1.2.2 Finding the constants
Given that the equation is of the form:
∞
X
f (x) = Cn ϕn (x)
n=1

we can multiply by ϕ∗m (x) and the weight function σ(x) and integrate over the length of the domain to
find: Z L Z L
∞
X
f (x)ϕ∗m (x)σ(x) dx = Cn ϕn (x)ϕ∗m (x)σ(x) dx
0 n=1 0
RL
We know that 0 ϕn (x)ϕ∗m (x)σ(x) dx = 0 if n ̸= m as the eigenfunctions are orthogonal with respect to
the weight function σ(x). So we conclude that:
RL
f (x)ϕ∗n (x)σ(x) dx
Cn = R0 L
0
|ϕn (x)|2 σ(x) dx

1.3 Eigenfunction expansion
Given an inhomogeneous PDE, such as:
ut = kuxx + Q(x, t)
we can solve it by using an eigenvalue expansion:
∞
X
u(x, t) = an (t)ϕn (x)
n=1

The eigenfunctions can be found by solving the equation without the term that makes it inhomogeneous
(Q(x, t) in this case). Then put the eigenfunction expansion back into the whole PDE and solve for the
inhomogeneous term. The inhomogeneous term should also be of the same form as the whole solution,
but with a different factor:
∞
X
Q(x, t) = qn (t)ϕn (x)
n=1
In general we can use orthogonality as:
RL
0
Q(x, t)ϕn (x) dx
qn (t) = RL
0
ϕ2n (x) dx
Please keep track of the boundaries of the variables. t > 0 for example gives:
Z t
∂y(t) ∂y(t)
= dt = y(t) − y(0)
∂t 0 ∂t

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