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Differential equations
Second order linear PDEs

Second order linear PDEs

Introduction

As we have already seen, in solving certain PDEs, we use methods used in solving ordinary differential equations to determine the solution of PDEs. In this section we discuss second order, linear PDEs with two independent variables: for transient problems these depend on time, tt and space, xx, while for steady-state problems, these may depend on a two-dimensional space system in xx and yy (Cartesian) or rr and θ\theta (polar). Examples of second order linear PDEs that we consider are the 1D wave, 1D diffusion and 2D Laplace equation given by Eqs. (2.78)-(2.80), respectively:

2yt2=c22yx2ut=a22ux22ux2+2uy2=0\begin{gathered} \frac{\partial^{2} y}{\partial t^{2}}=c^{2} \frac{\partial^{2} y}{\partial x^{2}} \\ \frac{\partial u}{\partial t}=a^{2} \frac{\partial^{2} u}{\partial x^{2}} \\ \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0 \end{gathered}

Before we move on to the general method of solving second order, linear PDEs we give some preliminaries on a class of ordinary differential equation (ODE) problems known as boundary-value problems (BVPs). Let us just note here that in Year I mathematics when you first encountered ODEs, the focus was on initial-value problems (IVPs) where initial conditions were given for the function (say, yy ) and its derivatives (y,y\left(y^{\prime}, y^{\prime \prime}\right., etc) at the same point in the independent variable, say x0x_{0}. Boundary value problems consist of a differential equation with an appropriate number of conditions describing the function and/or its derivatives. BVPs are generally more complicated than IVPs. For IVPs, and at least for linear ODEs, a unique, particular solution can be obtained.

Background: boundary-value problems

For BVPs the function and/or its derivatives are specified at different points of the independent variable; these are referred to as boundary conditions (BCs).

For example, consider the following ODE

y+p(x)y+q(x)y=f(x)y^{\prime \prime}+p(x) y^{\prime}+q(x) y=f(x)

where y=y(x)y=y(x). Since (2.81) is a second order ODE, we require two BCs for the BVP to be fully specified. Any of the following sets of BCs (or combinations of them) is appropriate for the solution of a BVP like the one given by (2.81):

(i) y(x0)=y0y\left(x_{0}\right)=y_{0} and y(x1)=y1y\left(x_{1}\right)=y_{1},

(ii) y(x0)=y0y^{\prime}\left(x_{0}\right)=y_{0} and y(x1)=y1y^{\prime}\left(x_{1}\right)=y_{1},

(iii) y(x0)=y0y\left(x_{0}\right)=y_{0} and y(x1)=y1y^{\prime}\left(x_{1}\right)=y_{1};

where y0,y1y_{0}, y_{1} are constant boundary values, x0,x1x_{0}, x_{1} are the boundary points and, x0x1x_{0} \neq x_{1}. Further, 'primes' denote differentiation with respect to xx. Unlike IVPs for which a unique solution was often guaranteed, depending on the constraints (i.e. BCs), BVPs might have:

(i) a unique solution, or

(ii) infinite solutions, or

(iii) no solution;

note that such behavior is easily possible even for the simplest, linear ODEs.

Solutions of BVPs

A function y(x)y(x) is a solution of a second order BVP if it

(i) is defined for all values of xx in the interval [x0,x1]\left[x_{0}, x_{1}\right],

(ii) is twice differentiable for all values of xx in the same interval,

(iii) satisfies the ODE [e.g. Eq. (2.81)],

(iv) satisfies the BCs at the two boundary points, say x0x_{0} and x1x_{1}.

Types of BCs

The solution of the BVP depends on the nature of the BCs. The three most common types of BCs\mathrm{BCs} that we will encounter are given as follows,

(i) y(x0)=y0y\left(x_{0}\right)=y_{0} and y(x1)=y1y\left(x_{1}\right)=y_{1}- Dirichlet BCs\mathrm{BCs}

(ii) y(x0)=y0y^{\prime}\left(x_{0}\right)=y_{0} and y(x1)=y1y^{\prime}\left(x_{1}\right)=y_{1}- Neumann BCs

(iii) y(x1)=y(x1)y\left(-x_{1}\right)=y\left(x_{1}\right) and y(x1)=y(x1)y^{\prime}\left(-x_{1}\right)=y^{\prime}\left(x_{1}\right)- periodic BCs

Note that Dirichlet BCs specify the function at given boundary points while Neumann BCs specify the derivative of the function at given boundary points. These can have physical meaning as we will see later in the course when dealing with the solution of certain PDEs.

Eigenvalues, eigenfunctions & homogeneous BVPs

Consider the general, second order, linear differential equation given by (2.81) with a set of BCs y0,y1y_{0}, y_{1}. An ODE is defined to be homogeneous if the forcing term, f(x)f(x) is zero. Linear ODEs with f(x)=0f(x)=0 have as a general solution:

y(x)=c1y1(x)+c2y2(x),y(x)=c_{1} y_{1}(x)+c_{2} y_{2}(x),

where c1c_{1} and c2c_{2} are arbitrary constants and y1(x)y_{1}(x) and y2(x)y_{2}(x) are linearly independent solutions to the ODE. The general solution (2.82) is a result of the principle of superposition. Now, a homogeneous BVP is one for which f(x)=0f(x)=0 and y0=y1=0y_{0}=y_{1}=0. If f(x)f(x) and/or any of the boundary values are nonzero, then the BVP is said to be inhomogeneous. More generally, we can define the homogeneity of a boundary condition using the following statement:

Definition 2.3 A boundary condition is homogeneous if y1(x)y_{1}(x) and

y2(x)y_{2}(x) satisfy the boundary condition and so does any linear combination, y(x)=c1y1(x)+c2y2(x)y(x)=c_{1} y_{1}(x)+c_{2} y_{2}(x)With Definition 2.3, it is easy to see that all types of boundary conditions given above are homogeneous (if y0=y1=0y_{0}=y_{1}=0 ) including the periodic BCs. We are interested in BVPs in which we have a differential equation for an unknown function, say y(x)y(x) and a real parameter, λ\lambda which will be considered as an undetermined constant that we need to determine. At this point, we will only discuss a very simple second order linear ODE, namely:

y+λy=0.y^{\prime \prime}+\lambda y=0 .

The topic falls into a larger class of ODE problems known as Sturm-Liouville problems (SLPs). The simplest type of such problems lead to half Fourier sine/cosine series or full Fourier series as shown in this section. More complicated SLPs may lead to Bessel functions and Hermite polynomials. The theory behind SLPs justify the method of separation of variables for linear second order ODEs which we discuss in Subsec. 2.4.

Here, our objective is to solve ODE (2.83) together with 2 boundary conditions (we will deal with each set of BCs given above, separately). Solutions of BVPs are related to eigenvalues and eigenvectors of linear equations: consider a square matrix A\mathbf{A} where we want to determine the values of λ\lambda that satisfy,

Ax=λx\mathbf{A} \underline{x}=\lambda \underline{x}

where λ\lambda are the eigenvalues, x\underline{x} are the eigenvectors with x0\underline{x} \neq 0 (i.e. we want nontrivial solutions). The same idea applies to BVPs: eigenvalues exist for nontrivial solutions or eigenfunctions, i.e. when y(x)0y(x) \neq 0. The rest of this section deals with finding all eigenvalues, λ\lambda for which ODE (2.83) has nontrivial solutions subject to the aforementioned three types of BCs.

Nontrivial solutions

As mentioned in the previous section, we consider Eq. (2.83) which, despite being very simple, turns out to be extremely useful and relevant in the solution of linear PDEs describing physical problems. Again, the ODE is given by:

y+λy=0,y^{\prime \prime}+\lambda y=0,

where the eigenvalues, λ\lambda can take any real value.

1. Homogeneous BVP with Dirichlet BCs (or fixed BCs)

Consider a set of homogeneous Dirichlet BCs on an arbitrary interval [0,L][0, L]

y(0)=y(L)=0,y(0)=y(L)=0,

where L>0L>0. In order to find all eigenvalues of the BVP (i.e. values of λ\lambda that give nontrivial solutions), we need to consider the following cases separately

λ>0,λ=0 and λ<0.\lambda>0, \quad \lambda=0 \text { and } \lambda<0 .

For λ>0\lambda>0, the general solution of Eq. (2.85) is given by

y(x)=c1cos(λx)+c2sin(λx).y(x)=c_{1} \cos (\sqrt{\lambda} x)+c_{2} \sin (\sqrt{\lambda} x) .

Applying BCs (2.86) in (2.87),

y(0)=c1=0c1=0,y(L)=c2sin(Lλ)=0.\begin{aligned} y(0) & =c_{1}=0 \quad \Rightarrow c_{1}=0, \\ y(L) & =c_{2} \sin (L \sqrt{\lambda})=0 . \end{aligned}

From Eq. (2.89),

c2=0 or sin(Lλ)=0c_{2}=0 \quad \text { or } \quad \sin (L \sqrt{\lambda})=0

since we are seeking nontrivial solutions, we need c20c_{2} \neq 0 hence we let c2c_{2} to be 'free' and choose

sin(Lλ)=0Lλ=nπ\sin (L \sqrt{\lambda})=0 \quad \Rightarrow \quad L \sqrt{\lambda}=n \pi

\therefore from Eq. (2.90), the eigenvalues are λn=(nπL)2\lambda_{n}=\left(\frac{n \pi}{L}\right)^{2}, where n1n \geq 1. The corresponding eigenfunctions are given by

yn(x)=c2sin(nπxL),y_{n}(x)=c_{2} \sin \left(\frac{n \pi x}{L}\right),

for any nonzero c2c_{2}. Equation (2.91) may be re-written in terms of a general coefficient, say bnb_{n},

yn(x)=bnsin(nπxL).y_{n}(x)=b_{n} \sin \left(\frac{n \pi x}{L}\right) .

For λ=0\lambda=0, the general solution of Eq. (2.85) is given by

y(x)=c3x+c4.y(x)=c_{3} x+c_{4} .

Applying BCs (2.86) in (2.93),

y(0)=c4=0c4=0,y(L)=c3L=0c3=0 since L>0.\begin{aligned} y(0) & =c_{4}=0 & \Rightarrow c_{4}=0, \\ y(L) & =c_{3} L=0 & \Rightarrow c_{3}=0 \text { since } L>0 . \end{aligned}

From Eqs. (2.94) and (2.95), c3=c4=0c_{3}=c_{4}=0 which, when substituted in (2.93), leads to the trivial solution y(x)=0y(x)=0 for all values of xx.

λ=0\Rightarrow \lambda=0 is not an eigenvalue of the BVP

For λ<0\lambda<0, the general solution of Eq. (2.85) is given by

y(x)=c5cosh(λx)+c6sinh(λx)y(x)=c_{5} \cosh (\sqrt{-\lambda} x)+c_{6} \sinh (\sqrt{-\lambda} x)

Applying BCs (2.86) in (2.96),

y(0)=c5=0c5=0,y(L)=c6sinh(Lλ)=0.\begin{aligned} & y(0)=c_{5}=0 \quad \Rightarrow c_{5}=0, \\ & y(L)=c_{6} \sinh (L \sqrt{-\lambda})=0 . \end{aligned}

We would choose c6c_{6} in Eq. (2.98) to be nonzero for nontrivial solutions. However, since sinh(Lλ)\sinh (L \sqrt{-\lambda}) can only be zero if its argument, LλL \sqrt{-\lambda} is zero and, given that L>0L>0 and λ<0\lambda<0, then Eq. (2.98) is only satisfied if c6=0c_{6}=0.

\therefore since c5=c6=0c_{5}=c_{6}=0, for λ<0\lambda<0, the only possible solution is the trivial one, y(x)=0y(x)=0.

λ<0\Rightarrow \lambda<0 are not eigenvalues of the BVP

2. Homogeneous BVP with Neumann BCs (or no-flux BCs)

Next, we consider homogeneous Neumann-type BCs on an arbitrary interval [0,L][0, L]

y(0)=y(L)=0,y^{\prime}(0)=y^{\prime}(L)=0,

where L>0L>0.

For the case of λ>0\lambda>0, applying (2.99) in (2.87), leads to

λn=(nπL)2,n1\lambda_{n}=\left(\frac{n \pi}{L}\right)^{2}, \quad n \geq 1

and the corresponding eigenfunctions are

yn(x)=ancos(nπxL).y_{n}(x)=a_{n} \cos \left(\frac{n \pi x}{L}\right) .

For the case of λ=0\lambda=0, applying (2.99) in (2.93), shows that λ=0\lambda=0 is an eigenvalue and the corresponding eigenfunction is

yn(x)=a0y_{n}(x)=a_{0}

where a0a_{0} is nonzero. As for Case 1 , it can be shown that the BVP does not have strictly negative eigenvalues.

\therefore for homogeneous BVPs with Neumann BCs on [0,L][0, L] all possible eigenvalues and corresponding eigenfunctions are given by

λn=(nπL)2,n0yn(x)=ancos(nπxL),n0\begin{aligned} \lambda_{n} & =\left(\frac{n \pi}{L}\right)^{2}, \quad n \geq 0 \\ y_{n}(x) & =a_{n} \cos \left(\frac{n \pi x}{L}\right), \quad n \geq 0 \end{aligned}

where n=0n=0 corresponds to the zero eigenvalue [Eq. (2.100)] and the corresponding eigenfunction [Eq. (2.101)].

3. Homogeneous BVP with periodic BCs

Lastly, we consider the following set of BCs on [L,L][-L, L]

y(L)=y(L),y(L)=y(L),y(-L)=y(L), \quad y^{\prime}(-L)=y^{\prime}(L),

where L>0L>0. Applying BCs (2.102) in (2.87) and (2.93), we obtain the following eigenvalues and corresponding eigenfunctions,

λn=(nπL)2,n0yn(x)=ancos(nπxL)+bnsin(nπxL),n0\begin{aligned} \lambda_{n} & =\left(\frac{n \pi}{L}\right)^{2}, \quad n \geq 0 \\ y_{n}(x) & =a_{n} \cos \left(\frac{n \pi x}{L}\right)+b_{n} \sin \left(\frac{n \pi x}{L}\right), \quad n \geq 0 \end{aligned}

where n=0n=0 corresponds to the zero eigenvalue [Eq. (2.103)] and the corresponding eigenfunction [Eq. (2.104)]. Again, it can be shown that this BVP does not have strictly negative eigenvalues. Example 2.3 Find all the nontrivial solutions (i.e. nonzero), y(x)y(x) of the following BVP:

y+y=0,y(0)=0,y(π)=1.y^{\prime \prime}+y=0, \quad y(0)=0, y^{\prime}(\pi)=1 .

Solution Here, the ODE is Eq. (2.85) with λ=1\lambda=1. We want to find whether nontrivial solutions exist for this specific value of λ\lambda. The general solution for Eq. (2.105) is:

y(x)=c1cosx+c2sinx,y(x)=c_{1} \cos x+c_{2} \sin x,

where c1c_{1} and c2c_{2} are arbitrary constants. Plugging the first boundary condition, y(0)=0y(0)=0 in Eq. (2.106) yields c1=0c_{1}=0. Differentiating Eq. (2.106) with respect to xx and applying the second boundary condition, y(π)=0y^{\prime}(\pi)=0 (with c1=0c_{1}=0 )

y(π)=c2cosπc2=1.y^{\prime}(\pi)=c_{2} \cos \pi \Rightarrow c_{2}=-1 .

In Eq. (2.106), this gives the solution as y(x)=sinxy(x)=-\sin x; note this is the only nontrivial solution subject to the boundary conditions.