Partial Differential Equations

PDEs are use to model many kinds of problems. Their solutions give evolution of a function $y(x,t)$ as a function of time and space. Boundary conditions involving time and space are used as initial conditions.

A method of separation of variables is used for solving them, where it is assumed that $V(x,y)=X(x)Y(y)$. Two other auxiliary ODE results are also needed:

$$\frac{d^{2}y}{d{x}^2}=a^{2}y⟹y=A_1{e}^{ax}+A_2{e}^{-ax}$$

$$\frac{d^{2}y}{d{x}^2}=-a^{2}y⟹y=B_1\sin \alpha x+B_2\cos ax$$

Another auxillary ODE are needed for some situations

$$\frac{dy}{dx}=-ay⟹y=C_1{e}^{-ax}$$

The general process for solving PDEs:

• Apply separation of variables
• Make an appropriate choice of constant $\pm {\mu }^2$
  • Nearly always $-{\mu }^2$
• Solve resulting ODEs
• Combine ODE solutions to form general PDE solution
• Apply boundary conditions to obtain particular PDE solution
  • Work out values for the arbitrary constants

Laplace's Equation

Laplace's equation described many problems involving flow in a plane:

$$\frac{{\partial }^{2}V(x,y)}{\partial x^2}+\frac{{\partial }^{2}V(x,y)}{\partial y^2}=0$$

Find the solution with the following boundary conditions:

• $V(0,y)=0$ and $V(\pi ,y)=0$
• $V(x,y)\to 0$ as $y\to \infty$
• $V(x,0)=10\sin x+7\sin 3x$

Starting with separation of variables:

$$\frac{\partial V(x,y)}{\partial x}=Y(y)\frac{dX(x)}{dx}\Rightarrow \frac{{\partial }^{2}V(x,y)}{\partial x^2}=Y(y)\frac{d^{2}X(x)}{d{x}^2}$$

$$\frac{\partial V(x,y)}{\partial y}=X(x)\frac{dY(y)}{dy}\Rightarrow \frac{{\partial }^{2}V(x,y)}{\partial y^2}=X(x)\frac{d^{2}Y(y)}{d{y}^2}$$

Substituting back into the original PDE:

$$Y(y)\frac{d^{2}X(x)}{d{x}^2}+X(x)\frac{d^{2}Y(y)}{d{y}^2}=0$$

$$\frac{1}{X(x)}\frac{d^{2}X(x)}{d{x}^2}=-\frac{1}{Y(y)}\frac{d^{2}Y(y)}{d{y}^2}$$

We have transformed the PDE into an ODE, where each side is a function of $x$/$y$ only. The only circumstances under which the two sides can be equal for all values of $x$ and $y$ is if both sides independent and equal to a constant. Since the constant is arbitrary, let it be ${\mu }^2$. Now we have two ODEs and their solutions from the auxiliary results earlier:

$$\frac{1}{X(x)}\frac{d^{2}X(x)}{d{x}^2}={\mu }^2-\frac{1}{Y(y)}\frac{d^{2}Y(y)}{d{y}^2}={\mu }^2$$

$$\frac{d^{2}X(x)}{d{x}^2}={\mu }^{2}X(x)\frac{d^{2}Y(y)}{d{y}^2}=-{\mu }^{2}Y(y)$$

$$X(x)=A_1{e}^{\mu x}+A_2{e}^{-\mu x}Y(y)=B_1\sin \mu y+B_2\cos \mu y$$

Substituting the solutions back into $V(x,y)=X(x)Y(y)$, we have a general solution to our PDE in terms of 4 arbitrary constants:

$$V(x,y)=(A_1{e}^{\mu x}+A_2{e}^{-\mu x})(B_1\sin \mu y+B_2\cos \mu y)$$

We can now apply boundary conditions:

• Substituting in $V(0,y)=0$ gives $A_1+A_2=0$
• Substituting in $V(\pi ,y)=0$ gives $A_1{e}^{\mu \pi }+A_2{e}^{-\mu \pi }=0$
• Using the two together gives $A_1(1-e^{-2\mu \pi )}=0$, so either:
  • $A_1=0$
  • $(1-e^{-2\mu \pi })=0$
• If $A_1=0$, then $A_2=0$, so $V(x,y)=0$
  • This is the trivial solution and is of no interest
• If $(1-e^{-2\mu \pi })=0$, then $\mu =0$
  • This also implies that $V(x,y)=0$, so is useless too

The issue is that we selected our arbitrary constant badly. If we use $-{\mu }^2$ instead, then our solutions are the other way round:

$$X(x)=A_1\sin \mu x+A_2\cos \mu xY(y)=B_1{e}^{\mu y}+B_2{e}^{-\mu y}$$

Checking the boundary conditions again:

• First condition, $V(0,y)=0$
  • Gives $A_2=0$
• Second condition $X(\pi ,y)=0$
  • Gives $A_1\sin \mu \pi =0$
    • Either $A_1=0$ (not interested)
    • $\sin \mu \pi =0$
  • $\mu$ is an integer, $n$

We now have:

$$V(x,y)=(A_1\sin nx)(B_1{e}^{ny}+B_2{e}^{-ny})$$

Where $n$ is any integer. Using the other boundary conditions:

• $V(x,y)\to 0$ as $y\to \infty$
• If $n$ is positive, then $B_1=0$ (otherwise $B_1{e}^{ny}\to \infty$)
• If $n$ is negative, then $B_2=0$ (otherwise $B_2{e}^{-ny}\to \infty$)

Taking $n$ as positive, the form of the solutions is:

$$V(x,y)=C_1{e}^{-y}\sin x(C_1=A_1{B}_2\text{for}n=1)$$

$$V(x,y)=C_2{e}^{-2y}\sin 2x(C_2=A_1{B}_2\text{for}n=2)$$

$$V(x,y)=C_n{e}^{-ny}\sin nx(C_n=A_1{B}_2\text{for}n)$$

The most general form is the sum of these:

$$\sum _{n=1}^{\infty }{C}_n{e}^{-ny}\sin nx$$

Applying the final boundary condition: $V(x,0)=10\sin x+7\sin 3x$

• $C_1=10$
• $C_3=7$
• $C_n=0$ for all other $n$

The complete solution is therefore:

$$V(x,y)=10e^{-y}\sin x+7e^{-3y}\sin 3x$$

The Heat Equation

The heat equation describe diffusion of energy or matter. With a diffusion coefficient $c$:

$$\frac{{\partial }^{2}u(x,t)}{\partial x^2}=\frac{1}{c^2}\frac{\partial u(x,t)}{\partial t}$$

Solving with the following boundary conditions:

• $u(0,t)=0$
• $u(2,t)=0$
• $u(x,0)=10$

Separating variables, $u(x,t)=X(x)T(t)$, and substituting, exactly the same as Laplace's equation, we have:

$$\frac{1}{X(x)}\frac{d^{2}X(x)}{d{x}^2}=\frac{1}{c^{2}T(t)}\frac{dT(t)}{dt}$$

Setting both sides again equal to a constant $-{\mu }^2$, we have two ODEs (one 2nd order, one 1st):

$$\frac{d^{2}X(x)}{d{x}^2}=-{\mu }^{2}X(x)\Rightarrow X(x)=A\cos \mu x+B\sin \mu x$$

$$\frac{dT(t)}{dt}=-{\mu }^2{c}^{2}T(t)\Rightarrow T(t)=C{e}^{-{\mu }^2{c}^{2}t}$$

The general solution is therefore:

$$u(x,y)=(A\cos \mu x+B\sin \mu x)C{e}^{-{\mu }^2{c}^{2}t}$$

Tidying up a bit, let $D=AC$, $E=BC$, $\lambda =\mu c$:

$$u(x,y)=(D\cos \frac{\lambda }{c}x+E\sin \frac{\lambda }{c}x)e^{-{\mu }^{2}t}$$

Applying the first boundary condition:

• $u(0,t)=0$
• Gives $0=D{e}^{-{\lambda }^{2}t}$
• Since $e^{-{\lambda }^{2}t}\ne 0$ for all $t$, $D=0$

We now have $u(x,t)=E\sin \left(\frac{\lambda x}{c}\right)e^{-{\lambda }^{2}t}$. The second boundary condition:

• $u(2,t)=0$, so $0=E\sin \left(\frac{2\lambda }{c}\right)e^{-{\lambda }^{2}t}$
• For the non trivial solution $E\ne 0$,and since $e^{-{\lambda }^{2}t}\ne 0$, $\sin \frac{2\lambda }{c}=0=n\pi$
• Therefore, $\lambda =\frac{n\pi c}{2}$ for $n=1,2,3,...$

Substituting this in gives:

$$u(x,t)=E\sin \left(\frac{n\pi x}{2}\right)e^{\frac{-n^2{\pi }^2{c}^{2}t}{4}}\text{for}n=1,2,3,...$$

The above equation is valid for any $n=1,2,3,...$, so summing these gives the most general solution:

$$u(x,t)=\sum _{n=1}^{\infty }{E}_n\sin \left(\frac{n\pi x}{2}\right)e^{\frac{-n^2{\pi }^2{c}^{2}t}{4}}$$

The last boundary condition is $u(x,0)=10$:

$$10=\sum _{n=1}^{\infty }{E}_n\sin \frac{n\pi x}{2}$$

This is in the form of the a Fourier series:

$$f(t)=\sum _{n=1}^{\infty }{b}_n\sin \frac{2\pi nt}{T}{b}_n=\frac{2}{T}{\int }_0^{T}f(t)\sin \frac{2\pi nt}{T}dt$$

We have:

$$10=\sum _{n=1}^{\infty }{E}_n\sin \frac{n\pi x}{2}{E}_n=\frac{2}{2}{\int }_0^{2}10\sin \frac{n\pi x}{2}dt$$

$${\int }_0^{2}10\sin \frac{n\pi x}{2}dt=10{\left[\frac{-2}{n\pi }\cos \frac{n\pi x}{2}\right]}_0^2=\frac{-20}{n\pi }[(-1{)}^n-1]$$

$$E_n=\{\begin{array}{ll}0& n\text{even}\\ \frac{40}{n\pi }& n\text{odd}\end{array}$$

Substituting this into $u$, and letting $n=2k-1$:

$$u(x,t)=\frac{40}{\pi }\sum _{n=1}^{\infty }\frac{1}{2k-1}\sin \left(\frac{(2k-1)\pi x}{2}\right)e^{-\frac{(2k-1{)}^2}{4}{\pi }^2{c}^{2}t}$$

The Wave Equation

The wave equation is used to describe vibrational problems:

$$\frac{{\partial }^{2}y(x,t)}{\partial x^2}=c^2\frac{{\partial }^{2}y(x,t)}{\partial t^2}$$

Solving the equation with the boundary conditions:

• $y(0,t)=0$
• $y(0,l)=0$
• $y(x,0)=\frac{\partial y(x,t)}{\partial t}$
• $y(x,0)=12\sin \frac{3\pi x}{l}$

Doing the usual separation of variables and substitution, and choosing a constant $-{\mu }^2$:

$$\frac{1}{X(x)}\frac{d^{2}X(x)}{d{x}^2}=\frac{c^2}{T(t)}\frac{d^{2}T(t)}{d{t}^2}=-{\mu }^2$$

Solving both ODEs:

$$\frac{d^{2}X(x)}{d{x}^2}=-{\mu }^{2}X(x)\Rightarrow X(x)=A\cos \mu x+B\sin \mu x$$

$$\frac{d^{2}T(x)}{d{t}^2}=\frac{-{\mu }^2}{c^2}T(t)\Rightarrow T(t)=C\cos \frac{\mu }{c}t+D\sin \frac{\mu }{c}t$$

$$y(x,t)=(A\cos \mu x+B\sin \mu x)(C\cos \frac{\mu }{c}t+D\sin \frac{\mu }{c}t)$$

This is the general solution. Start applying boundary conditions:

• $y(0,t)=0$ implies that $0=A(C\cos \frac{\mu }{c}t+D\sin \frac{\mu }{c}t)$
  • As this is true for all $t$, $A=0$
• $y(0,t)=l$ implies that $0=B\sin \mu l(C\cos \frac{\mu }{c}t+D\sin \frac{\mu }{c}t)$
  • This is also true for all $t$, so $B\sin \mu l=0$
  • Required that $B!=0$, so $0=\sin \mu l$
  • $\mu =\frac{n\pi }{l}$ for $n=1,2,3,...$

We now have:

$$y(x,t)=B\sin \frac{n\pi }{l}x(C\cos \frac{\mu }{c}t+D\sin \frac{\mu }{c}t)\text{for}n=1,2,3,...$$

Applying the third boundary condition, $y(x,0)=\frac{\partial y(x,t)}{\partial t}$:

$$\frac{\partial y(x,t)}{\partial t}=B\sin \frac{n\pi }{l}x(-C\frac{\mu }{c}\sin \frac{\mu }{c}t+D\frac{\mu }{c}\cos \frac{\mu }{c}t)$$

$$\frac{\partial y(x,t)}{\partial t}=0=B\sin \frac{n\pi }{l}x(D\frac{\mu }{c})$$

As this is for all $x$, $B\sin \frac{n\pi }{l}x\ne 0$, so $D=0$. We now have:

$$y(x,t)=E\sin \frac{n\pi x}{l}\cos \frac{n\pi t}{cl}\text{for}n=1,2,3,...$$

The general solution is then:

$$y(x,t)=\sum _{n=1}^{\infty }{E}_n\sin \frac{n\pi x}{l}\cos \frac{n\pi t}{cl}$$

Applying the final boundary condition of $y(x,0)=12\sin \frac{3\pi x}{l}$, gives $E_3=12$, else $E_n=0$. The particular solution is therefore:

$$y(x,t)=12\sin \frac{3\pi x}{l}\cos \frac{3\pi t}{cl}$$