A partial differential equation (PDE) is a relation involving one or more functions of several variables, and their partial derivatives. The order of a partial differential equation is the order of the highest partial derivative that appears in the equation.

There are three classical partial differential equations of order two which are very common in physical applications. They are known as the heat equation (1), the wave equation (2), and the Laplace equation (3).

Heat Equation

Consider the boundary-value problem

First, it’s easy to see that the equation is linear, which means any linear combination of solutions $u_1\left(x,t\right),\dots,u_n\left(x,t\right)$ is again a solution. This suggests the following “plan” for solving this boundary-value problem.

1. Find as many solutions $u_i\left(x,t\right)$ as we can that satisfies the boundary condition $u\left(0,t\right)=u\left(l,t\right)=0$.

2. Find a proper linear combination of $u_i$ to satisfies the initial condition $u\left(x,0\right)=f\left(x\right)$.

An important method to solve partial differential equations is separation of variables, which means we first assume $u\left(x,t\right)=X\left(x\right)T\left(t\right)$. Then

Plug this in the original equation and get

divide both side by $\alpha^2XT$ gives

Notice that the left-hand side only depend on $x$, while the right-hand side only depend on $t$. The only way that they can equal for any $x$ and $t$ is that both of them are constant. Thus,

By separation of variable, we have transformed a PDE into two solvable ODEs,

Now, plug in the boundary condition.

Since $T\left(t\right)\ne 0$ (not identically zero, otherwise $u$ would be identically zero, and we have a trivial solution), we have $X\left(0\right)=X\left(l\right)=0$.

Now, we only need to solve

and

The first one is a second-order, linear, constant coefficient ODE, which we can directly write out the solution. There are three different cases:

• $\lambda = 0$: The general solution is $X\left(x\right)=c_1x+c_2$ for some constant $c_1, c_2$. Plug in the boundary condition, we have $c_1=c_2=0$. Thus, there are no non-trivial solution for $\lambda = 0$.

• $\lambda<0$: The general solution is $X\left(x\right)=c_1e^{\sqrt{-\lambda}x}+c_1e^{\sqrt{-\lambda}x}$ for some constant $c_1,c_2$. The boundary condition implies

This linear system of equation has non-trivial solution $c_1, c_2$ if and only if

This implies $e^{2\sqrt{-\lambda}l}=1$, which is impossible since $2\sqrt{-\lambda}l>0$. Therefore, there are also no non-trivial solution for $\lambda < 0$.

• $\lambda > 0$: The general solution is $X\left(x\right)=c_1\cos\sqrt{\lambda}x+c_2\sin\sqrt{\lambda}x$ for some constant $c_1,c_2$ The boundary condition implies

For $c_2\ne0$, we need to have $\sqrt{\lambda}l=n\pi$ for positive integer $n$, or $\displaystyle \lambda_n=\frac{n^2\pi^2}{l^2}$. Then the equation has non-trivial solutions

Then, we have the second equation,

which has solution

Hence, we have

Finally, we need to use the initial condition to find a proper linear combination of $u_n$’s, i.e., find coefficients $c_n$’s such that

satisfies $u\left(x,0\right)=f\left(x\right)$. This means that

It’s easily seen that $c_n$’s should be the Fourier coefficient of the pure sine expansion, which is

Finally,

is the desired solution.

Wave Equation

Consider the boundary-value problem

This problem can also be solved by separation of variable. Let $u\left(x,t\right)=X\left(x\right)T\left(t\right)$, then

Plug into the original equation, we have

or two ODEs,

and

We’ve seen in the previous problem that the first equation has solutions

For the second equation, it becomes

The solutions are

Thus,

is a non-trivial solution of the BVP for every positive integer $n$, and every suitable constants $a_n, b_n$. The linear combination

satisfies

To satisfy the initial condition, we only need to have

By Fourier sine expansion, we have