# What is the relation between separation of variables and the eigenfunctions and eigenvalues for PDEs?

Studying Fourier Series and its application of solutions for Partial Differential Equations, in particular (historically) for the heat equation, one starts by separating variables. Somehow related to this is the concept of eigenfunctions and eigenvalues, which give a solution of the derivative operator.

I have not found a simple reading on the relationship between these two concepts. Can somebody explain in "intuitive" language, what is the connection between separation of variables, eigenfunctions and eigenvalues for the differential operator, and Fourier Solutions for a PDE problem?

If you take the heat equation with zero temperature and finite ends

$$\frac{\partial u}{\partial t} =k\frac{\partial^{2} u}{\partial x^{2}} \tag{1}$$

$$0 < x < L , t> 0$$

$$u(0,t) = u(L,t) = 0 \tag{2}$$

this is called a homogeneous Dirichlet Boundary problem. Our initial condition is given as

$$u(x,0) = f(x) \tag{3}$$

when we separate them

## Separation

$$u(x,t) = \phi(x) G(t) \tag{4}$$

We're saying that $$u(x,t)$$ is formed from a linear combination of the eigenfunctions $$\phi(x)$$ and $$G(t)$$ which we need to solve the eigenvalues for. We get the eigenvalues from the boundary conditions.

$$\frac{\partial u}{\partial t} = \phi(x) \frac{\partial G}{\partial t} \tag{5}$$ $$\frac{\partial^{2} u}{\partial x^{2}} = \frac{d \phi^{2}}{dx^{2}} G(t) \tag{6}$$

$$\phi(x) \frac{dG}{dt} = k \frac{d^{2} \phi}{dx^{2}} G(t) \tag{7}$$ $$\frac{1}{G} \frac{dG}{dt} = k\frac{1}{\phi} \frac{d^{2}\phi}{dx^{2}} \tag{8}$$

Which gives us

$$\frac{d^{2}\phi}{dx^{2}} =-\lambda \tag{9}$$ $$\frac{dG}{dt} = -\lambda k G \tag{10}$$

Solving this with the conditons we get

$$G(t) = ce^{-\lambda kt} \tag{11}$$

if $$\lambda > 0$$ $$\phi(x) = c_{1} \cos(\sqrt{\lambda} x) + c_{2} \sin(\sqrt{\lambda} x) \tag{12}$$

in which case we put in our boundary conditions

$$\phi(0) = 0 \implies c_{1} = 0 \tag{13}$$

now we know that $$\phi(x) = \sin(\sqrt{\lambda}x)$$

$$\phi(L) = c_{2}\sin(\sqrt{\lambda}L) = 0 \tag{14}$$ we solve this for the eigenvalues. Sine is zero multiples of $$n\pi$$ however there is the $$L$$

$$\sqrt{\lambda}L = n\pi \implies \lambda_{n} = (\frac{n \pi }{L})^{2} \tag{15}$$

## Putting series together

$$u(x,t) = \sum_{n=1}^{\infty} B_{n} \sin( \sqrt{\lambda} x) e^{ -k \lambda t } \tag{16}$$

$$B_{n} = \frac{2}{L} \int_{0}^{L} f(x) \sin(\sqrt{\lambda} x) dx \tag{17}$$

Change boundaries to Neumann? $$\frac{d\phi}{dx}(0) = 0 \tag{17}$$ $$\frac{d\phi}{dx}(L) = 0 \tag{18}$$ $$u(x,t) = \sum_{n=0}^{\infty} A_{n} \cos( \sqrt{\lambda} x) e^{ -k \lambda t } \tag{17}$$ $$A_{0} = \frac{1}{L} \int_{0}^{L} f(x) dx \tag{19}$$ $$A_{n} = \frac{2}{L} \int_{0}^{L} f(x) \cos(\sqrt{\lambda}x) dx \tag{20}$$ Change to mixed $$\phi(-L) = \phi(L) \tag{21}$$

$$\frac{d\phi}{dx}(-L) = \frac{d\phi}{dx}(L) \tag{22}$$

$$u(x,t) = \sum_{n=0}^{\infty} A_{n} \cos( \sqrt{\lambda} x) e^{ -k \lambda t } +\sum_{n=1}^{\infty} B_{n} \sin(\sqrt{\lambda} x) e^{ -k \lambda t } \tag{23}$$ i.e.

same idea

## Method of Eigenfunction Expansion

In general we can do something else. Suppose we have the heat equation with non-homogeneous boundaries and time dependent source.

$$\frac{\partial u}{\partial t} = k\frac{\partial^{2} u}{\partial x^{2}} + Q(x,t) \tag{24}$$

with boundaries

$$u(0,t) = A(t) \\ u(L,t) = B(t) \tag{25}$$

$$u(x,0) = f(x) \tag{26}$$

we first homogeneize

$$\frac{\partial v}{\partial t} = k\frac{\partial^{2} v}{\partial x^{2} } + \hat{Q}(x,t) \tag{27}$$

$$v(0,t) = 0 \\ v(L,t) = 0 \\ v(x,0) = g(x) \tag{28}$$

giving us

$$\frac{d^{2}\phi }{dx^{2}} +\lambda \phi =0 \\ \phi(0) = 0 \\ \phi(L) = 0 \tag{29}$$

then we know This is called the eigenfunction expansion $$v(x,t) = \sum_{n=1}^{\infty} a_{n}(t) \phi_{n}(x) \tag{30}$$ In order to meet the initial condition we have

$$g(x) = \sum_{n=1}^{\infty} a_{n}(0) \phi_{n}(x) \tag{31}$$

to get the coefficients then

$$a_{n}(0) =\frac{\int_{0}^{L} g(x) \phi_{n}(x) dx}{\int_{0}^{L} \phi_{n}^{2}(x) dx}$$

## In order to solve the rest of the problem

$$\frac{\partial v}{\partial t} = \sum_{n=1}^{\infty} \frac{da_{n}(t)}{dt}\phi_{n}(x) \tag{32}$$

$$\frac{\partial^{2} v}{\partial x^{2}} = -\sum_{n=1}^{\infty} -a_{n}(t) \lambda_{n}\phi_{n}(x) \tag{33}$$

if we subsitute this into $$27$$

$$\frac{da_{n}}{dt} + \lambda_{n}ka_{n} = \frac{\int_{0}^{L} \hat{Q}(x,t) \phi_{n}(x) dx}{ \int_{0}^{L} \phi_{n}^{2}(x) dx } = \hat{q}_{n}(t)\tag{34}$$

this gives us an eigenfunction expansion for $$\hat{Q}(x,t)$$

$$\hat{Q}(x,t) = \sum_{n=1}^{\infty} \hat{q}_{n}(t) \phi_{n}(x)$$

you have to solve this by the following way

$$a_{n}(t) = a_{n}(0) e^{-\lambda_{n} kt} + e^{ \lambda_{n} kt } \int_{0}^{t} \hat{q}_{n}(\tau) e^{\lambda_{n} k \tau} d \tau \tag{35}$$

which gives us our solution

$$v(x,t) = \sum_{n=1}^{\infty} a_{n}(t) \phi_{n}(x) \tag{36} \\ \textrm{where } a_{n}(t) = a_{n}(0)e^{-\lambda_{n} k t}$$

• This looks like a great walkthrough into solving the heat equation. However I am curious about a generalization of the concepts used in here. How are in general eigenfunctions and its eigenvalues used for solving PDEs? Is it intrinsically connected to separation of variables? – hirschme Oct 1 '18 at 19:22
• I will update this in a little bit to address that. It's lunch time. – Shogun Oct 1 '18 at 19:30