Differential equations – Partial differential equations – Laplace equation

The two-dimensional heat conduction and wave equation are respectively:

\[\alpha^2\left(u_{xx}+u_{yy}\right)=u_t\quad\quad\text{and}\quad\quad a^2\left(u_{xx}+u_{yy}\right)=u_{tt}\quad\quad\text{with}\quad u=u(x,y,t).\]

Three-dimensional this is:

\[\alpha^2\left(u_{xx}+u_{yy}+u{zz}\right)=u_t\quad\quad\text{and}\quad\quad a^2\left(u_{xx}+u_{yy}+u_{zz}\right)=u_{tt}\quad\quad\text{met}\quad u=u(x,y,z,t).\]

We restrict ourselves to the stable situation, where there occurs no change in the time, that is: \(u_t=0\). In both cases (heat conduction and wave equation) we then have:

\[u_{xx}+u_{yy}=0\quad\text{(2-dimensional)}\quad\quad\text{or}\quad\quad u_{xx}+u_{yy}+u_{zz}=0\quad\text{(3-dimensional)}.\]

This is called the Laplace equation or potential equation.

Now consider for instance a Dirichlet problem for a rectangle:

\[\left\{\begin{array}{l}u_{xx}+u_{yy}=0,\quad 0 < x < a,\quad 0 < y < b\\[2.5mm] u(x,0)=f_1(x),\quad u(x,b)=f_2(x),\quad 0 < x < a\\[2.5mm] u(0,y)=g_1(y),\quad u(a,y)=g_2(y),\quad 0 < y < b.\end{array}\right.\]

In the case of a heat conduction problem the function \(u=u(x,y)\) then describes the temperature at the point \((x,y)\) in a thin metal plate, which thickness might be neglected, with length \(a\) and width \(b\). The metal plate is perfectly isolated through the full surface. At the boundaries there is a temperature distribution attached described by the functions \(f_1(x)\), \(f_2(x)\), \(g_1(y)\) and \(g_2(y)\).

Instead one can also consider a Neumann problem for a rectangle:

\[\left\{\begin{array}{l}u_{xx}+u_{yy}=0,\quad 0 < x < a,\quad 0 < y < b\\[2.5mm] u_y(x,0)=f_1(x),\quad u_y(x,b)=f_2(x),\quad 0 < x < a\\[2.5mm] u_x(0,y)=g_1(y),\quad u_x(a,y)=g_2(y),\quad 0 < y < b.\end{array}\right.\]

Combinations of a Dirichlet and a Neumann problem are also possible; for instance if one side is isolated and there are heat sources attached to the other boundaries.

Dirichlet problem for a rectangle

Now consider the Dirichlet problem

\[\left\{\begin{array}{l}u_{xx}+u_{yy}=0,\quad 0 < x < a,\quad 0 < y < b\\[2.5mm] u(x,0)=f(x),\quad u(x,b)=0,\quad 0 < x < a\\[2.5mm] u(0,y)=0,\quad u(a,y)=0,\quad 0 < y < b\end{array}\right.\]

with three homogeneous boundary conditions and one nonhomogeneous boundary condition.

Again we use the method of separation of variables: let \(u(x,y)=X(x)Y(y)\), then we have:

\[u_{xx}+u_{yy}=0\quad\Longleftrightarrow\quad X''(x)Y(y)+X(x)Y''(y)=0.\]

For \(u(x,y)=X(x)Y(y)\neq 0\) we now have (divide by \(X(x)Y(y)\)): \(\displaystyle\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}=0\;\Longleftrightarrow\;\frac{X''(x)}{X(x)}=-\frac{Y''(y)}{Y(y)}\). The left-hand side only depends on \(x\) (and not on \(y\)), while the right-hand side only depends on \(y\) (and not on \(x\)). Therefore they can only be equal for all \(x\in(0,a)\) and \y\in(0,b)\) if they are constant. Hence:

\[\frac{X''(x)}{X(x)}=-\frac{Y''(y)}{Y(y)}=\sigma\quad(\text{separation constant}).\]

This implies:

\[X''(x)-\sigma X(x)=0\quad\text{and}\quad Y''(y)+\sigma Y(y)=0.\]

Using the homogeneous boundary conditions we now have (with \(X(x)\neq 0\) and \(Y(y)\neq 0\)):

\[u(x,b)=0:\quad X(x)Y(b)=0\quad\Longrightarrow\quad Y(b)=0,\] \[u(0,y)=0:\quad X(0)Y(y)=0\quad\Longrightarrow\quad X(0)=0\]

and

\[u(a,y)=0:\quad X(a)Y(y)=0\quad\Longrightarrow\quad X(a)=0.\]

So for \(X(x)\) we obtain the homogeneous boundary-value problem:

\[\left\{\begin{array}{l}X''(x)-\sigma X(x)=0,\quad 0 < x < a\\[2.5mm]X(0)=0,\quad X(a)=0.\end{array}\right.\]

This is the same homogeneous boundary-value problem that we encountered earlier at the heat conduction equation and the wave equation with eigenvalues \(\sigma_n=-\displaystyle\frac{n^2\pi^2}{a^2}\) and eigenfunctions \(X_n(x)=\displaystyle\sin\left(\frac{n\pi x}{a}\right)\) with \(n=1,2,3,\ldots\). For \(Y(y)\) we now obtain: \(Y_n''(y)-\displaystyle\frac{n^2\pi^2}{a^2}Y_n(y)=0\) with \(Y_n(b)=0\) with eigenfunctions \(Y_n(y)=\displaystyle\sinh\left(\frac{n\pi(b-y)}{a}\right),\;n=1,2,3,\ldots\). Hence:

\[u_n(x,y)=X_n(x)Y_n(y)=\sin\left(\frac{n\pi x}{a}\right)\sinh\left(\frac{n\pi(b-y)}{a}\right),\quad n=1,2,3,\ldots.\]

Using the super position principle we then have:

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

Now we use the nonhomogeneous boundary condition:

\[u(x,0)=f(x)\quad\Longleftrightarrow\quad\sum_{n=1}^{\infty}\sinh\left(\frac{n\pi b}{a}\right)c_n\sin\left(\frac{n\pi x}{a}\right)=f(x).\]

This is a Fourier sine series for \(f(x)\). Using the Euler-Fourier formulas we then have:

\[\sinh\left(\frac{n\pi b}{a}\right)c_n=\frac{2}{a}\int_0^af(x)\sin\left(\frac{n\pi x}{a}\right)\,dx,\quad n=1,2,3,\ldots.\]

A more general Dirichlet problem for a rectangle can now be split into such problems with three homogeneous boundary conditions. These can be solved in a similar way and the sum of these solutions then is the solution of the general Dirichlet problem.

Dirichlet problem for a circle

For a circular domain we might use polar coordinates: \(\left\{\begin{array}{l}x=r\cos(\theta)\\[2.5mm]y=r\sin(\theta)\end{array}\right.\) with \(r\geq 0\) and \(0\leq\theta < 2\pi\).

Then the Laplace equation \(u_{xx}+u_{yy}=0\) transfers into: \(\displaystyle u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}=0\).

A Dirichlet problem for a circle with radius \(R>0\) then is:

\[\left\{\begin{array}{l}u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}=0,\quad 0 < r < R,\quad 0 < \theta < 2\pi\\[2.5mm] u(R,\theta)=f(\theta),\quad 0 \leq\theta < 2\pi.\end{array}\right.\]

Again we use the method of separation of variables: let \(u(r,\theta)=R(r)T(\theta)\), then we have:

\[u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}=0\quad\Longleftrightarrow\quad R''(r)T(\theta)+\frac{1}{r}R'(r)T(\theta)+\frac{1}{r^2}R(r)T''(\theta)=0.\]

For \(u(r,\theta)=R(r)T(\theta)\neq 0\) we then have (divide by \(R(r)T(\theta)\)):

\[\frac{R''(r)}{R(r)}+\frac{1}{r}\frac{R'(r)}{R(r)}+\frac{1}{r^2}\frac{T''(\theta)}{T(\theta)}=0\quad\Longleftrightarrow\quad r^2\frac{R''(r)}{R(r)}+r\frac{R'(r)}{R(r)}=-\frac{T''(\theta)}{T(\theta)}=\sigma\quad(\text{separation constant}).\]

This implies:

\[r^2R''(r)+rR'(r)-\sigma R(r)=0\quad\text{and}\quad T''(\theta)+\sigma T(\theta)=0.\]

Here we have some (hidden) boundary conditions: the function \(T(\theta)\) should be \(2\pi\)-periodic and the function \(R(r)\) should be bounded for \(r\downarrow 0\).

We distinguish three possibilities: \(\sigma=0\), \(\sigma < 0\) and \(\sigma > 0\):

• \(\sigma=0\): \(T''(\theta)=0\;\Longrightarrow\;T(\theta)=a_1\theta+a_2\). This is only periodic if \(a_1=0\). Hence: \(\sigma_0=0\) is an eigenvalue with eigenfunction \(T_0(\theta)=1\).


• \(\sigma=-\mu^2 < 0\): \(T''(\theta)-\mu^2T(\theta)=0\;\Longrightarrow\;T(\theta)=b_1\cosh(\mu\theta)+b_2\sinh(\mu\theta)\). This is only periodic if \(b_1=b_2=0\) (the trivial solution).


• \(\sigma=\mu^2 > 0\): \(T''(\theta)+\mu^2T(\theta)=0\;\Longrightarrow\;T(\theta)=c_1\cos(\mu\theta)+c_2\sin(\mu\theta)\). This is only \(2\pi\)-periodic if \(\mu=\pm n\) with \(n\in\{1,2,3,\ldots\}\). This leads to the eigenvalues (\sigma_n=n^2,\;n=1,2,3,\ldots\) and the eigenfunctions \(T_n(\theta)=c_n\cos(n\theta)+k_n\sin(n\theta),\;n=1,2,3,\ldots\).

The differential equation for \(R(r)\) is an Euler differential equation.

For \(\sigma=0\) we have: \(r^2R''(r)+rR'(r)=0\) with general solution \(R(r)=c_1+c_2\ln(r)\). Since \(R(r)\) should be bounded for \(r\downarrow 0\), we conclude that \(c_2=0\). So an eigenfunction is \(R_0(r)=1\).

For \(\sigma=n^2\) we have: \(r^2R''(r)+rR'(r)-n^2R(r)=0\) with general solution \(R(r)=k_1r^n+k_2r^{-n}\). Since \(R(r)\) should be bounded for \(r\downarrow 0\) we conclude that \(k_2=0\). So eigenfunctions are \(R_n(r)=r^n\) met \(n=1,2,3,\ldots\).

Hence:

\[u_n(r,\theta)=R_n(r)T_n(\theta)=c_nr^n\cos(n\theta)+k_nr^n\sin(n\theta),\quad n=0,1,2,\ldots.\]

Using the super position principle we now have:

\[u(r,\theta)=\frac{c_0}{2}+\sum_{n=1}^{\infty}r^n\left(c_n\cos(n\theta)+k_n\sin(n\theta)\right).\]

The boundary condition \(r(R,\theta)=f(\theta)\) for \(0\leq\theta < 2\pi\) then leads to:

\[\frac{c_0}{2}+\sum_{n=1}^{\infty}R^n\left(c_n\cos(n\theta)+k_n\sin(n\theta)\right)=f(\theta).\]

This is an ordinary Fourier series for \(f(\theta)\). Note that \(f(\theta)\) is defined on the interval \([0,2\pi)\) and that this can be extended as a periodic function with period \(2\pi\). Using the Euler-Fourier formulas we then have: \(c_0=\displaystyle\frac{1}{\pi}\int_0^{2\pi}f(\theta)\,d\theta\),

\[R^nc_n=\frac{1}{\pi}\int_0^{2\pi}f(\theta)\cos(n\theta)\,d\theta\quad\text{and}\quad R^nk_n=\frac{1}{\pi}\int_0^{2\pi}f(\theta)\sin(n\theta)\,d\theta,\quad n=1,2,3,\ldots.\]

This implies that we have obtained the solution:

\[u(r,\theta)=\frac{c_0}{2}+\sum_{n=1}^{\infty}r^n\left(c_n\cos(n\theta)+k_n\sin(n\theta)\right).\]

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Last modified on August 23, 2021