MA 30300: Differential Equations

9.7. Steady-State Temperature and Laplace’s Equation

Goal: Understand how to solve boundary value problems for Laplace’s equation using separation of variables and Fourier series.

0. Motivation

We study the temperature distribution in a thin two-dimensional plate occupying a region \(R\) in the \(xy\)-plane. Assuming the plate is insulated and thin. The temperature function satisfies the two-dimensional heat equation:

\[ \frac{\partial u}{\partial t} = k \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) = k \nabla^2 u \]

In steady state (where the temperature does not vary with time), this reduces to Laplace’s equation:

\[ \nabla^2 u = 0 \]

1. Rectangular Domain (Finite Plate)

Rectangular Domain: Solve the boundary value problem

\[ u_{xx} + u_{yy} = 0, \quad 0 < x < a,\; 0 < y < b \] \[ u(x,0) = f_1(x),\quad u(x,b) = f_2(x),\quad u(0,y) = g_1(y),\quad u(a,y) = g_2(y) \]

Remark: This problem can be split into four problems, each with a single nonhomogeneous boundary condition.

Example: Consider the boundary value problem

\[ u_{xx} + u_{yy} = 0 \] \[ u(x,0) = f(x),\quad u(x,b)=0,\quad u(0,y)=0,\quad u(a,y)=0 \]

Show that the solution is

\[ u(x,y) = \sum_{n=1}^{\infty} c_n \sin\left(\frac{n\pi x}{a}\right) \sinh\left(\frac{n\pi (b-y)}{a}\right) \]
where
\[ c_n= \frac{2}{a\,\sinh\left(\frac{n\pi b}{a}\right)} \int_0^a f(x)\sin\left(\frac{n\pi x}{a}\right)\,dx \]

Solution (Separation of Variables): Substituting \( u(x,y) = X(x)Y(y) \) into Laplace’s equation gives

\[ X''(x)Y(y) + X(x)Y''(y) = 0 \] \[ \frac{X''}{X} = - \frac{Y''}{Y} = -\lambda \]

This leads to the system

\[ X'' + \lambda X = 0, \quad X(0)=X(a)=0 \] \[ Y'' - \lambda Y = 0, \quad Y(b)=0. \]

The eigenvalue problem for \(X\) gives

\[ \lambda_n = \frac{n^2\pi^2}{a^2}, \quad X_n(x) = \sin\left(\frac{n\pi x}{a}\right), \quad n=1,2,3,\dots \]

Then \(Y\) satisfies

\[ Y'' - \frac{n^2\pi^2}{a^2} Y = 0 \]

whose general solution is

\[ Y_n(y) = A_n \cosh\left(\frac{n\pi y}{a}\right) + B_n \sinh\left(\frac{n\pi y}{a}\right) \]

Applying the boundary condition \(Y_n(b)=0\) gives

\[ Y_n(y) = c_n \sinh\left(\frac{n\pi (b-y)}{a}\right) \]

Thus the solution is

\[ u(x,y) = \sum_{n=1}^{\infty} c_n \sin\left(\frac{n\pi x}{a}\right) \sinh\left(\frac{n\pi (b-y)}{a}\right) \]

To satisfy \(u(x,0)=f(x)\), we require

\[ \sum_{n=1}^{\infty} c_n \sinh\left(\frac{n\pi b}{a}\right) \sin\left(\frac{n\pi x}{a}\right) = f(x) \]

Thus, using Fourier sine series,

\[ c_n \sinh\left(\frac{n\pi b}{a}\right) = \frac{2}{a} \int_0^a f(x)\sin\left(\frac{n\pi x}{a}\right)\,dx \]

so the coefficients are

\[ c_n = \frac{2}{a\,\sinh\left(\frac{n\pi b}{a}\right)} \int_0^a f(x)\sin\left(\frac{n\pi x}{a}\right)\,dx \]

2. Semi-Infinite Strip

Semi-Infinite Strip:

\[ u_{xx} + u_{yy} = 0 \] \[ u(x,0) = 0,\quad u(x,b) = 0 \quad (0 < x < \infty) \] \[ u(0,y) = g(y), \quad u(x,y)\ \text{is bounded as } x \to \infty \]

Solution (Separation of Variables): Assume

\[ u(x,y) = X(x)Y(y) \]

Substituting into \(u_{xx} + u_{yy} = 0\) gives

\[ X''(x)Y(y) + X(x)Y''(y) = 0 \] \[ \frac{Y''}{Y} = -\frac{X''}{X} = - \lambda \]

Thus we obtain the two ODEs

\[ Y'' + \lambda Y = 0, \quad Y(0)=0,\; Y(b)=0 \] \[ X'' - \lambda X = 0 \]

First solve the eigenvalue problem in \(y\). This gives

\[ \lambda_n = \frac{n^2\pi^2}{b^2}, \quad Y_n(y) = \sin\left(\frac{n\pi y}{b}\right), \quad n=1,2,3,\dots \]

Next solve the equation for \(X_n(x)\):

\[ X_n'' - \frac{n^2\pi^2}{b^2} X_n = 0 \]

The general solution is

\[ X_n(x) = A_n e^{\frac{n\pi x}{b}} + B_n e^{-\frac{n\pi x}{b}} \]

Since \(u(x,y)\) must remain bounded as \(x \to \infty\), we must set \(A_n = 0\). Hence

\[ X_n(x) = B_n e^{-\frac{n\pi x}{b}} \]

Therefore

\[ u_n(x,y) = B_n e^{-\frac{n\pi x}{b}} \sin\left(\frac{n\pi y}{b}\right) \]

We get the formal solution

\[ u(x,y) = \sum_{n=1}^{\infty} b_n e^{-\frac{n\pi x}{b}} \sin\left(\frac{n\pi y}{b}\right) \]

Using the boundary condition \(u(0,y)=g(y)\), we obtain the Fourier sine series

\[ g(y) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi y}{b}\right) \]

Thus the coefficients are

\[ b_n = \frac{2}{b} \int_0^b g(y)\sin\left(\frac{n\pi y}{b}\right)\,dy \]

Final solution:

\[ u(x,y) = \sum_{n=1}^{\infty} b_n e^{-\frac{n\pi x}{b}} \sin\left(\frac{n\pi y}{b}\right) \]
where
\[ b_n = \frac{2}{b} \int_0^b g(y)\sin\left(\frac{n\pi y}{b}\right)\,dy \]

3. Dirichlet Problem for a Circular Disk

Dirichlet Problem for a Circular Disk:

We now investigate the steady-state temperature in a circular disk of radius \(a\) with insulated faces and given boundary temperatures. To match the geometry of the disk, we express in polar coordinates

\[ x = r\cos\theta, \quad y = r\sin\theta, \quad \text{and} \quad u(x,y)=u(r,\theta) \]
where
\[ r = \sqrt{x^2+y^2} \ (0 < r < a), \quad \tan(\theta) = \frac{y}{x} \]
We rewrite Laplace’s equation
\[ u_{xx} + u_{yy} = 0 \]
in polar coordinates.
\[ u_x = u_r \frac{\partial r}{\partial x} + u_\theta \frac{\partial \theta}{\partial x}, \quad u_y = u_r \frac{\partial r}{\partial y} + u_\theta \frac{\partial \theta}{\partial y}. \]

Since \(u\) is now a function of \(r\) and \(\theta\), we use the chain rule to express derivatives in terms of \(r\) and \(\theta\). First compute the first derivatives:

\[ \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2+y^2}} = \frac{x}{r} = \cos\theta \]
\[ \frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2+y^2}} = \frac{y}{r} = \sin\theta \]
\[ \frac{\partial \theta}{\partial x} = -\frac{y}{x^2+y^2} = -\frac{\sin\theta}{r} \]
\[ \frac{\partial \theta}{\partial y} = \frac{x}{x^2+y^2} = \frac{\cos\theta}{r} \]

Substituting these into the first derivatives gives:

\[ u_x = u_r \cos\theta - \frac{1}{r}u_\theta \sin\theta, \quad u_y = u_r \sin\theta + \frac{1}{r}u_\theta \cos\theta. \]

Now we differentiate again and combine \(u_{xx}+u_{yy}\). After simplification we obtain:

\[ u_{xx} + u_{yy} = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta}. \]

The Laplace’s equation \(u_{xx}+u_{yy}=0\) becomes the polar form:

\[ u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} = 0 \]

with \(0 < r < a\), with boundary condition

\[ u(a,\theta) = f(\theta), \quad \forall \theta \]
Here the function \(f(\theta)\) of period \(2\pi\) is given. We solve for \(u(r,\theta)\), which is also of period \(2\pi\)
\[ u(r,\theta) = u(r,\theta + 2\pi). \]

Summary (Boundary Value Problem): We consider Laplace’s equation in polar coordinates:

\[ u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} = 0 \]

for \(0 < r < a\), with boundary condition

\[ u(a,\theta) = f(\theta), \quad \forall \theta, \quad \text{ (the function \(f(\theta)\) of period \(2\pi\) is given) } \]

and periodicity condition

\[ u(r,\theta) = u(r,\theta + 2\pi) \]

Solution (Separation of Variables)

Assume

\[ u(r,\theta) = R(r)\Theta(\theta) \]
\[ u(r,\theta)=R(r)\Theta(\theta) \]

Compute derivatives:

\[ u_r = R'(r)\Theta(\theta), \quad u_{rr} = R''(r)\Theta(\theta) \]
\[ u_\theta = R(r)\Theta'(\theta), \quad u_{\theta\theta} = R(r)\Theta''(\theta) \]
\[ R''\Theta + \frac{1}{r}R'\Theta + \frac{1}{r^2}R\Theta'' = 0 \]

Multiply by \(r^2\):

\[ r^2 R''\Theta + r R'\Theta + R\Theta'' = 0 \]

Divide by \(R\Theta\):

\[ \frac{r^2 R''}{R} + \frac{r R'}{R} + \frac{\Theta''}{\Theta} = 0 \]

Rearrange:

\[ \frac{r^2 R'' + r R'}{R} = -\frac{\Theta''}{\Theta} = \lambda \]

This yields two ODEs:

\[ \Theta'' + \lambda \Theta = 0 \] \[ r^2 R'' + r R' - \lambda R = 0 \]

We solve the angular equation

\[ \Theta'' + \lambda \Theta = 0, \quad \Theta(\theta)=\Theta(\theta+2\pi) \]

Case 1: \(\lambda = 0\).

\[ \Theta'' = 0 \] \[ \Theta' = C_1 \] \[ \Theta(\theta) = C_1 \theta + C_2 \]

Periodicity condition \(\Theta(\theta)=\Theta(\theta+2\pi)\) gives

\[ C_1 \theta + C_2 = C_1(\theta+2\pi) + C_2 \Rightarrow C_1=0 \]
Therefore
\[ \Theta(\theta) = C_2 \]

Since eigenfunctions are defined up to a constant multiple, we can normalize the constant and choose \(C_2 = 1\), so that

\[ \Theta_0(\theta)=1 \]

Case 2: \(\lambda > 0\). Write \(\lambda = n^2\)

\[ \Theta'' + n^2 \Theta = 0 \] \[ \Theta(\theta) = A \cos(n\theta) + B \sin(n\theta) \]

Apply periodicity \(\Theta(\theta)=\Theta(\theta+2\pi)\):

\[ \cos(n(\theta+2\pi))=\cos(n\theta+2n\pi)=\cos(n\theta) \] \[ \sin(n(\theta+2\pi))=\sin(n\theta+2n\pi)=\sin(n\theta) \]

This holds only if

\[ n \in \mathbb{Z} \]

Thus eigenvalues are

\[ \lambda_n = n^2, \quad n=0,1,2,\dots \]

and eigenfunctions are

\[ \Theta_0(\theta)=1, \quad \Theta_n(\theta)=A_n \cos(n\theta)+B_n \sin(n\theta) \]

To ensure periodicity \(\Theta(\theta)=\Theta(\theta+2\pi)\), we obtain eigenvalues

\[ \lambda_n = n^2, \quad n=0,1,2,\dots \]

with eigenfunctions

\[ \Theta_0 = 1,\quad \Theta_n(\theta)=A_n\cos(n\theta)+B_n\sin(n\theta) \]

Case 3: \(\lambda < 0\). Let

\[ \lambda = -\alpha^2, \quad \alpha > 0 \]

Then

\[ \Theta'' - \alpha^2 \Theta = 0 \]

So the general solution is

\[ \Theta(\theta) = A e^{\alpha \theta} + B e^{-\alpha \theta}, \quad \alpha > 0 \]

Apply periodicity condition:

\[ \Theta(\theta) = \Theta(\theta + 2\pi) \]

Substitute:

\[ A e^{\alpha \theta} + B e^{-\alpha \theta} = A e^{\alpha(\theta+2\pi)} + B e^{-\alpha(\theta+2\pi)} \]

Rearrange:

\[ A e^{\alpha \theta}(1 - e^{2\pi \alpha}) + B e^{-\alpha \theta}(1 - e^{-2\pi \alpha}) = 0 \]

Since this holds for all \(\theta\), first set \(\theta = 0\):

\[ A(1 - e^{2\pi \alpha}) + B(1 - e^{-2\pi \alpha}) = 0 \quad (1) \]

Now set \(\theta = 1\):

\[ A e^{\alpha}(1 - e^{2\pi \alpha}) + B e^{-\alpha}(1 - e^{-2\pi \alpha}) = 0 \quad (2) \]

We now have a linear system in \(A\) and \(B\):

\[ \begin{cases} A(1 - e^{2\pi \alpha}) + B(1 - e^{-2\pi \alpha}) = 0 \\ A e^{\alpha}(1 - e^{2\pi \alpha}) + B e^{-\alpha}(1 - e^{-2\pi \alpha}) = 0 \end{cases} \]

The only solution is

\[ A = 0, \quad B = 0. \]

So there are no nontrivial periodic solutions for \(\lambda < 0\).

Summary: the angular eigenfunctions are:

\[ \Theta_0(\theta)=1, \quad \lambda_0 = 0 \]
\[ \Theta_n(\theta)=A_n\cos(n\theta)+B_n\sin(n\theta), \quad n=1,2,3,\dots, \quad \lambda_n = n^2 \]

We now solve the radial equation for each case:

\[ r^2 R'' + rR' - \lambda R = 0 \]

Case 1: \(\lambda = 0\)

\[ r^2 R_0'' + rR_0' = 0 \]

Divide by \(r\) (for \(r>0\)):

\[ rR_0'' + R_0' = 0 \]

Recognize a product rule:

\[ \frac{d}{dr}(rR_0') = R_0' + rR_0'' \]

Thus the equation becomes:

\[ \frac{d}{dr}(rR_0') = 0 \]
\[ rR_0' = C \] \[ R_0' = \frac{C}{r} \] \[ R_0(r) = C_0 + C_1 \ln r \]

We want \(u(r,\theta)\), hence \(R_0(r)\), to be continuous at \(r=0\), so \(C_1 = 0\), and therefore

\[ R_0(r) = C_0. \]

Case 2: \(\lambda = n^2,\; n \ge 1\)

\[ r^2 R'' + rR' - n^2 R = 0 \]

Try \(R(r)=r^k\):

\[ R' = kr^{k-1}, \quad R'' = k(k-1)r^{k-2} \]
\[ r^2 k(k-1)r^{k-2} + r k r^{k-1} - n^2 r^k = 0 \] \[ k(k-1) + k - n^2 = 0 \] \[ k^2 - n^2 = 0 \] \[ k = \pm n \]
\[ R_n(r)=C_n r^n + D_n r^{-n} \]

We want \(u(r,\theta)\), hence \(R_n(r)\), to be continuous at \(r=0\).

\[ R_n(r)=C_n r^n + D_n r^{-n} \]

Since \(r^{-n} \to \infty\) as \(r \to 0\) (for \(n \ge 1\)), we must have

\[ D_n = 0 \]

Therefore

\[ R_n(r)=C_n r^n. \]

Combining the radial and angular solutions, we obtain the formal series solution

\[ u(r,\theta) = R_0(r)\Theta_0(\theta) + \sum_{n=1}^{\infty} R_n(r)\Theta_n(\theta) \]
\[ = \frac{a_0}{2} + \sum_{n=1}^{\infty} r^n \big( a_n \cos(n\theta) + b_n \sin(n\theta) \big) \]

To satisfy the boundary condition

\[ u(a,\theta) = f(\theta), \]

we evaluate at \(r=a\):

\[ f(\theta) = u(a,\theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a^n \big( a_n \cos(n\theta) + b_n \sin(n\theta) \big) \]

This must match the Fourier series of \(f(\theta)\). The coefficients are

\[ \frac{a_0}{2}=\frac{1}{\pi}\int_0^{2\pi} f(\theta)\,d\theta \] \[ a_n=\frac{1}{\pi a^n}\int_0^{2\pi} f(\theta)\cos(n\theta)\,d\theta \] \[ b_n=\frac{1}{\pi a^n}\int_0^{2\pi} f(\theta)\sin(n\theta)\,d\theta \]

Final solution:

\[ u(r,\theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty} r^n \big( a_n \cos(n\theta) + b_n \sin(n\theta) \big) \]

where

\[ \frac{a_0}{2}=\frac{1}{\pi}\int_0^{2\pi} f(\theta)\,d\theta \] \[ a_n=\frac{1}{\pi a^n}\int_0^{2\pi} f(\theta)\cos(n\theta)\,d\theta \] \[ b_n=\frac{1}{\pi a^n}\int_0^{2\pi} f(\theta)\sin(n\theta)\,d\theta \]

4. Exercises

Exercise 1 (Dirichlet Problem on a Rectangle): Solve the boundary value problem

\[ u_{xx} + u_{yy} = 0, \quad 0 < x < a,\; 0 < y < b \] \[ u(x,0) = 0, \quad u(x,b) = 0 \] \[ u(0,y) = 0, \quad u(a,y) = g(y) \]

Use the method of separation of variables to find the solution.

Answer:

\[ u(x,y) = \sum_{n=1}^{\infty} c_n \frac{\sinh\left(\frac{n\pi x}{b}\right)} {\sinh\left(\frac{n\pi a}{b}\right)} \sin\left(\frac{n\pi y}{b}\right), \]

where

\[ c_n = \frac{2}{b} \int_0^b g(y)\sin\left(\frac{n\pi y}{b}\right)\,dy. \]

Exercise 2 (Dirichlet Problem on a Rectangle): Solve the boundary value problem

\[ u_{xx} + u_{yy} = 0, \quad 0 < x < a,\; 0 < y < b \] \[ u(x,0) = 0, \quad u(0,y) = 0, \quad u(a,y) = 0 \] \[ u(x,b) = f(x) \]

Answer:

\[ u(x,y) = \sum_{n=1}^{\infty} c_n \frac{\sinh\left(\frac{n\pi y}{a}\right)} {\sinh\left(\frac{n\pi b}{a}\right)} \sin\left(\frac{n\pi x}{a}\right), \]

where

\[ c_n = \frac{2}{a} \int_0^a f(x)\sin\left(\frac{n\pi x}{a}\right)\,dx. \]

Exercise 3: Find a solution of Laplace’s equation in the rectangle \(0 < x < a,\; 0 < y < b\)

\[ u_{xx} + u_{yy} = 0 \] \[ u_y(x,0) = 0,\quad u_y(x,b) = 0 \] \[ u(a,y) = 0,\quad u(0,y) = g(y) \]

Answer:

\[ u(x,y) = \frac{c_0 (a - x)}{2a} + \sum_{n=1}^{\infty} c_n \, \frac{\sinh\!\left(\frac{n\pi (a - x)}{b}\right)} {\sinh\!\left(\frac{n\pi a}{b}\right)} \cos\!\left(\frac{n\pi y}{b}\right) \] \[ c_n = \frac{2}{b} \int_0^b g(y)\cos\!\left(\frac{n\pi y}{b}\right)\,dy, \quad n = 1,2,3,\dots \] \[ c_0 = \frac{2}{b} \int_0^b g(y)\,dy \]

Exercise 4: Find a solution to Laplace’s equation in the semi-infinite strip \(0 < x < a,\; y > 0\), subject to the boundary conditions

\[ u_{xx} + u_{yy} = 0 \] \[ u(0,y) = 0, \quad u(a,y) = 0, \quad u(x,0) = f(x) \] \[ \text{and } u(x,y) \text{ is bounded as } y \to \infty. \]

Answer:

\[ u(x,y) = \sum_{n=1}^{\infty} c_n \, e^{-\frac{n\pi y}{a}} \sin\!\left(\frac{n\pi x}{a}\right) \] where \[ c_n = \frac{2}{a} \int_0^a f(x)\sin\!\left(\frac{n\pi x}{a}\right)\,dx \]

Exercise 5:

The solution to the following Laplace's equation

\[ u_{xx}+u_{yy}=0, \quad 0 < x < 2,\; 0 < y < 3 \] \[ u(x,0)=0, \quad u(x,3)=0 \] \[ u_x(0,y)=0, \quad u(2,y)=4 \]

can be written as

\[ u(x,y)=\sum_{n=1}^{\infty} c_n u_n \]

where \(c_n\) are constants and \(u_n\) are the fundamental solutions. Find \(u_n\).

  1. \[ u_n=\cosh \left(\dfrac{n\pi y}{3}\right)\sin \left(\dfrac{n\pi x}{3}\right) \]
  2. \[ u_n=\sinh \left(\dfrac{n\pi y}{2}\right)\sin \left(\dfrac{n\pi x}{2}\right) \]
  3. \[ u_n=\cosh \left(\dfrac{n\pi x}{2}\right)\sin \left(\dfrac{n\pi y}{2}\right) \]
  4. \[ u_n=\sinh \left(\dfrac{n\pi x}{3}\right)\sin \left(\dfrac{n\pi y}{3}\right) \]
  5. \[ u_n=\cosh \left(\dfrac{n\pi x}{3}\right)\sin \left(\dfrac{n\pi y}{3}\right) \]

Solution: We use separation of variables \(u(x,y)=X(x)Y(y)\). From the boundary conditions \(u(x,0)=u(x,3)=0\), we obtain the eigenvalue problem

\[ Y''+\lambda Y=0, \quad Y(0)=0,\; Y(3)=0 \]

which gives

\[ \lambda_n=\left(\frac{n\pi}{3}\right)^2,\quad Y_n(y)=\sin\left(\frac{n\pi y}{3}\right) \]

For \(X(x)\), we solve

\[ X''-\left(\frac{n\pi}{3}\right)^2 X=0 \]

so

\[ X_n(x)=A_n\cosh\left(\frac{n\pi x}{3}\right)+B_n\sinh\left(\frac{n\pi x}{3}\right) \]

Using \(u_x(0,y)=0\), we get \(B_n=0\), hence

\[ X_n(x)=\cosh\left(\frac{n\pi x}{3}\right) \]

Therefore

\[ u_n(x,y)=\cosh\left(\frac{n\pi x}{3}\right)\sin\left(\frac{n\pi y}{3}\right). \]

Exercise 6: Consider Dirichlet’s problem for the region exterior to the circle \(r = a\).

\[ r^2 u_{rr} + r u_r + u_{\theta\theta} = 0 \] \[ u(a,\theta) = f(\theta), \quad \text{and } u(r,\theta) \text{ is bounded as } r \to \infty. \]
Derive the solution
\[ u(r,\theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \frac{1}{r^n} \left( a_n \cos(n\theta) + b_n \sin(n\theta) \right) \]
where
\[ a_n = \frac{a^n}{\pi} \int_0^{2\pi} f(\theta) \cos(n\theta) \, d\theta, \quad (n = 0, 1, 2, \dots) \] \[ b_n = \frac{a^n}{\pi} \int_0^{2\pi} f(\theta) \sin(n\theta) \, d\theta, \quad (n = 1, 2, 3, \dots) \]