Preface
This section demonstrates application of the resolvent method for solving Laplace's equation in the first quadrant with mixed boundary conditions.
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Introduction to Linear Algebra with Mathematica
Glossary
Resolvent method for Laplace's equation
𝔉 ℱ
We demonstrate application of the resolvent method for solving Laplace's equation in the first quadrant
\[
\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 , \qquad x> 0, \quad y> 0,
\]
under the mixed boundary conditions:
\[
u(0,y) - \left. \frac{\partial u}{\partial x} \right\vert_{x=0} = f(y) , \qquad
u(x,0) - \left. \frac{\partial u}{\partial y} \right\vert_{y=0} = 0 .
\]
Since the domain ℝ²+ is unbounded, we need to impose the regularity conditions at infinity that requires the unknown function u(x, y) and its derivative to approach zero at infinity. Also, the boundary of quadrant is not smooth at the corner, we need to impose the wedge condition on the function u(x, y), so it is assumed to be bounded near the origin. The given function f(y) is assumed to be integrable and satisfy the regularity condition at y = 0, so f(0) = 0.
In order to solve the above problem, we break it into two auxiliary problems, from which we construct the required solution. It is convenient to introduce the projection operators:
\begin{align*}
\left[ B_1 u(x,y) \right] (y) &= u(0,y) = \lim_{x\to 0} \, u(x,y) ,
\\
\left[ B_2 u(x,y) \right] (y) &= u_x (0,y) = \lim_{x\to 0} \, \frac{\partial u}{\partial x} .
\end{align*}
Then the above boundary condition at x = 0 can be written in compact form:
\[
B_1 u - B_2 u = f .
\]
First auxiliary boundary value problem
\[
\begin{split}
\Delta\,u(x,y) = 0 , \qquad (x,y) \in \mathbb{R}^2_{+} ,
\\
u(0,y) = g(y) , \qquad
u(x,0) - \left. \frac{\partial u}{\partial y} \right\vert_{y=0} = 0 .
\end{split}
\]
To solve the first auxiliary problem, we apply sine-Fourier transform with respect to independent variable x. Correspondingly, we denote by v its sine-Fourier transform:
\[
v(y) = ℱ_s\left[ u(x,y) \right] = u_s^F \left( k, y \right) = \int_0^{\infty} u(x,y)\,\sin \left( k\,x \right) {\text d} x .
\]
Then using the formula
\[
ℱ_s \left[ \frac{{\text d}^2 f}{{\text d} x^2} \right] (k) =
\int_0^{\infty} f'' (x)\,\sin (kx) \,{\text d}x = k\, f (0) - k^2
ℱ_s \left[ f \right] (k) ,
\]
we transfer the the first auxiliary boundary value problem to the following problem with respect to its sine Fourier transform v:
\[
\frac{{\text d}^2 v}{{\text d} y^2} - k^2 v (y) = - k\,g(y) , \qquad \left( v - \frac{{\text d} v}{{\text d} y} \right)_{y=0} = 0 .
\]
To solve the above problem, we apply variation of parameters method. So we seek its solution in the form
\[
v(y) = A\, e^{-ky} + B\, e^{ky} ,
\]
where coefficients A and B are assumed to be functions of y. Their derivatives are solutions of the algebraic system of equations
\[
\begin{split}
A' \,e^{-ky} + B'\, e^{ky} &= 0 ,
\\
-k\,A' \,e^{-ky} + k\,B'\, e^{ky} &= -k\,g(y) .
\end{split}
\]
It is not hard to solve the above system of algebraic equations to obtain
\[
A' = \frac{1}{2}\, e^{ky}\, g(y) , \qquad B' = - \frac{1}{2}\, e^{-ky}\, g(y) .
\]
Integrating yields
\[
A(y) = \frac{1}{2}\, \int_0^{y} {\text d} \xi \,e^{k\xi}\, g(\xi ) , \qquad B(y) = \frac{1}{2}\, \int_y^{\infty} {\text d} \xi \,e^{-k\xi}\, g(\xi ) ,
\]
where we dropped constant of integration that we will incorporate shortly. Using these expressions, we find the general solution that satisfies the regularity condition at infinity to be
\[
v(y) = \frac{1}{2}\,\int_0^{\infty} {\text d} \xi \,e^{-k|y - \xi |}\, g(\xi )
+C\, e^{-ky},
\]
where C is an arbitrary constant. To satisfy the boundary condition at y = 0, we substitute the above function into this condition to obtain
\[
\frac{1-k}{2}\, \int_0^{\infty} {\text d} \xi \,e^{-k\xi}\, g(\xi ) + C \left( 1 + k \right) = 0 .
\]
The above equation can be solved with respect to C to obtain its value and the solution becomes
\[
v(y) = \frac{1}{2}\,\int_0^{\infty} {\text d} \xi \,e^{-k|y- \xi |}\, g(\xi ) -
\frac{1}{2}\,\frac{1-k}{1+k} \, \int_0^{\infty} {\text d} \xi \,e^{-k \left( y+\xi \right)}\, g(\xi ) .
\]
Applying the inverse sine Fourier transform and using the identities
\begin{align*}
,\frac{1-k}{1+k} &= -1 + \frac{2}{1+k} ,
\\
\int_0^{\infty} \sin \left( kx \right) e^{-kA}\,{\text d} k &= \frac{x}{A^2 + x^2} , \qquad A = |y-\xi |,
\end{align*}
Assuming[x > 0 && A > 0,
Integrate[Sin[k*x]*Exp[-A*k], {k, 0, Infinity}]]
x/(x^2 + y^2)
we obtain
\begin{align*}
u(x,y) &= ℱ_s^{-1} \left[ v(y) \right] = \frac{2}{\pi} \int_0^{\infty} v(y)\,\sin \left( kx \right) {\text d} k
\\
&= \frac{x}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{g(\xi )}{(y-\xi )^2 + x^2} + \frac{x}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{g(\xi )}{(y+\xi )^2 + x^2} - \frac{2}{\pi} \int_0^{\infty} {\text d} \xi \, g(\xi ) \int_0^{\infty} \frac{{\text d}k}{1+k} \,e^{-k \left( y + \xi \right)} \,\sin \left( k\xi \right) .
\end{align*}
The first integral can be modified as
\begin{align*}
u(x,y) &= \frac{x}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{g(\xi ) - g(y)}{(y-\xi )^2 + x^2} + \frac{g(y)}{\pi} \, \int_0^{\infty} {\text d} \xi \,\frac{x}{(y-\xi )^2 + x^2}
\\
& \quad +
\frac{x}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{g(\xi )}{(y+\xi )^2 + x^2} - \frac{2}{\pi} \int_0^{\infty} {\text d} \xi \, g(\xi ) \int_0^{\infty} \frac{{\text d}k}{1+k} \,e^{-k \left( y + \xi \right)} \,\sin \left( k\xi \right) .
\end{align*}
The second integral is easy to evaluate
\[
\int_0^{\infty} {\text d} \xi \,\frac{x}{(y-\xi )^2 + x^2} = \pi
\]
because
\[
\int_0^{\infty} {\text d} \xi \,\frac{x}{(y-\xi )^2 + x^2} = \lim_{\varepsilon \to 0} \left[ \int_0^{y-\varepsilon} \frac{x}{(y-\xi )^2 + x^2}\, {\text d} \xi + \int_{y+\varepsilon}^{\infty} \frac{x}{(y-\xi )^2 + x^2}\, {\text d} \xi \right] = \pi .
\]
We denote the operator in the right-hand side by S1:
\[
S_1 \left[ g(y) \right] (x,y) = \frac{x}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{g(\xi ) - g(y)}{(\xi - y)^2 + x^2} + g(y ) -
\frac{x}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{g(\xi )}{(\xi + y)^2 + x^2} .
\]
We can similarly break the last integral into two parts:
\[
\frac{x}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{g(\xi )}{(\xi + y)^2 + x^2} = \frac{x}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{g(\xi ) - g(0)}{(\xi + y)^2 + x^2} + g(0) \,\frac{1}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{x}{(\xi + y)^2 + x^2}
\]
Integration yields
Assuming[0 < y && x > 0,
Integrate[x/((y + s)^2 + x^2), {s, 0, Infinity}]]
1/2 (\[Pi] - 2 ArcTan[y/x])
\[
\frac{1}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{x}{(\xi + y)^2 + x^2} = \frac{1}{2} - \frac{1}{\pi}\, \arctan \left(
\frac{y}{x} \right) .
\]
Therefore, we get the solution to the first auxiliary problem:
\[
S_1 \left[ g(y) \right] (x,y) = g(y ) - \frac{g(0)}{2} + \frac{g(0)}{\pi}\,\arctan \left(
\frac{y}{x} \right) - g(0) \,\frac{2}{\pi} \int_0^{\infty} \frac{{\text d}k}{1+k}\, \frac{\sin kx}{k}\, e^{-ky} + \frac{x}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{g(\xi ) - g(y)}{(\xi - y)^2 + x^2} -
\frac{x}{\pi} \int_0^{\infty} {\text d} \xi \,\frac{g(\xi ) - g(0)}{(\xi + y)^2 + x^2} - \frac{2}{\pi} \int_0^{\infty} {\text d}\xi \left[ g(\xi ) - g(0) \right] \int_0^{\infty} \frac{{\text d}k}{1+k}\, e^{-k(y+ \xi )} \sin (kx) .
\]
Since \( \lim_{x\to 0} \,\arctan \left(
\frac{y}{x} \right) = \pi /2 \) for y > 0, we have
\[
B_1 S_1 g = \lim_{x\to 0} \,S_1 \left[ g \right] =
\begin{cases}
g(y) , & \ \mbox{ for } y > 0 ,
\\
\frac{1}{2}\, g(0) , & \ \mbox{ for } y=0.
\end{cases}
\]
Now we need to find the limit
\[
B_2 S_1 g = \lim_{x\to 0} \, \frac{\partial}{\partial x} \,S_1 \left[ g \right] = \frac{1}{\pi}\,\lim_{x\to 0} \, \frac{\partial}{\partial x}\,x\, \,\int_{0}^{\infty} \frac{g(\xi ) - g(y)}{\left( \xi - y \right)^2 + x^2} \,{\text d} \xi -
\frac{1}{\pi}\,\lim_{x\to 0} \, \frac{\partial}{\partial x}\,x\, \,\int_{0}^{\infty} \frac{g(\xi ) - g(0)}{\left( \xi + y \right)^2 + x^2} \,{\text d} \xi + \frac{g(0)}{\pi}\,\lim_{x\to 0} \, \frac{y}{y^2 + x^2} .
\]
Each of the integrals
\[
I_1 = \frac{1}{\pi}\,\lim_{x\to 0} \,\frac{\partial}{\partial x}\,x\, \,\int_{0}^{\infty} \frac{g(\xi ) - g(y)}{\left( \xi - y \right)^2 + x^2} \,{\text d} \xi ,
\]
and
\[
I_2 = \frac{1}{\pi}\,\lim_{x\to 0} \,\frac{\partial}{\partial x}\,x\, \,\int_{0}^{\infty} \frac{g(\xi )- g(0)}{\left( \xi + y \right)^2 + x^2} \,{\text d} \xi .
\]
can be evaluated using integration by parts.
For example, consider the integral I1, which we break into two parts, based on Mathematica help.
Simplify[D[ArcTan[t/x], t]]
x/(t^2 + x^2)
D[x/(t^2 + x^2), t]
-((2 t x)/(t^2 + x^2)^2)
D[x/(t^2 + x^2), x]
-((2 x^2)/(t^2 + x^2)^2) + 1/(t^2 + x^2)
\[
I_1 = \frac{1}{\pi}\,\lim_{x\to 0} \, \frac{\partial}{\partial x}\,x\, \,\int_{0}^{\infty} \frac{g(\xi ) - g(y)}{\left( \xi - y \right)^2 + x^2} \,{\text d} \xi =
I_{1a} + I_{1b} ,
\]
where
\[
I_{1a} = \frac{1}{\pi}\,\lim_{x\to 0} \,\int_{0}^{\infty} \frac{\partial}{\partial \xi} \,\frac{g(\xi ) - g(y)}{\left( \xi - y \right)^2 + x^2} \,{\text d} \xi
\]
and
\[
I_{1b} = \frac{1}{\pi}\,\lim_{x\to 0} \,\int_{0}^{\infty}
\frac{g(\xi ) - g(y)}{\left( \xi - y \right)^2 + x^2} \,{\text d} \xi =
\frac{1}{\pi}\,\int_{0}^{\infty} \frac{g(\xi ) - g(y)}{\left( \xi - y \right)^2} \,{\text d} \xi .
\]
Integration by parts yields
\begin{align*}
I_{1a} &= x\,\int_{0}^{\infty} \frac{g(\xi ) - g(y)}{\xi - y}\,\frac{\partial}{\partial \xi} \left( \frac{x}{(\xi - y)^2 + x^2} \right)
\\
&= \lim_{\varepsilon \to 0} x\,\int_{0}^{y-\varepsilon} \frac{g(\xi ) - g(y)}{\xi - y}\,\frac{\partial}{\partial \xi} \left( \frac{x}{(\xi - y)^2 + x^2} \right) + \lim_{\varepsilon \to 0} x\,\int_{y+\varepsilon}^{\infty} \frac{g(\xi ) - g(y)}{\xi - y}\,\frac{\partial}{\partial \xi} \left( \frac{x}{(\xi - y)^2 + x^2} \right)
\\
&= \lim_{\varepsilon \to 0} x\left[ \frac{g(y-\varepsilon ) - g(y)}{-\varepsilon} - \frac{g(y+\varepsilon ) - g(y)}{\varepsilon} \right] - x \,\frac{g(0) - g(y)}{-y}
\\
& \quad - x\,\int_{0}^{\infty} \frac{x}{(\xi - y)^2 + x^2} \frac{\partial}{\partial \xi} \left( \frac{g(\xi ) - g(y)}{\xi - y} \right)
\end{align*}
So when taking the limit, we get
\[
\lim_{x\to 0} I_{1a} (x) = 0 ,
\]
subject that g(y) has two derivatives (which is assumed).
Finally, we represent the limit as
\[
B_2 S_1 g = \lim_{x\to 0} \, \frac{\partial}{\partial x} \,S_1 \left[ g \right] = \frac{1}{\pi}\,\int_{0}^{\infty} \frac{g(\xi ) - g(y)}{\left( \xi - y \right)^2} \,{\text d} \xi - \frac{1}{\pi}\,\int_{0}^{\infty} \frac{g(\xi ) - g(0)}{\left( \xi + y \right)^2} \,{\text d} \xi + \frac{g(0)}{\pi\,y} .
\]
Second auxiliary boundary value problem
\[
\begin{split}
\Delta\,u(x,y) = 0 , \qquad (x,y) \in \mathbb{R}^2_{+} ,
\\
\left. \frac{\partial u}{\partial x} \right\vert_{x=0} = g(y) , \qquad
u(x,0) - \left. \frac{\partial u}{\partial y} \right\vert_{y=0} = 0 .
\end{split}
\]
To solve the above problem, we apply cosine-Fourier transform with respect to independent variable x:
\[
v(y) = ℱ_c \left[ u \right] = u_c^F \left( k, y \right) = \int_0^{\infty} u(x,y)\,\cos \left( k\,x \right) {\text d} x .
\]
Using the known relation (that is established upon integration by parts),
\[
ℱ_c \left[ \frac{{\text d}^2 f}{{\text d} x^2} \right] (k) =
\int_0^{\infty} f'' (x)\,\cos (kx) \,{\text d}x = - f' (0) - k^2
ℱ_c \left[ f \right] (k) ,
\]
we transfer the the second auxiliary boundary value problem to the following problem with respect to its cosine Fourier transform:
\[
\frac{{\text d}^2 v}{{\text d} y^2} - k^2 v (y) = g(y) , \qquad \left( v - \frac{{\text d} v}{{\text d} y} \right)_{y=0} = 0 .
\]
To solve the above boundary value problem, we use variation of parameters, and represent the required solution as
\[
y(y) = A\,e^{-ky} + B\,e^{ky} .
\]
To satisfy the inhomogeneous differential equation, we set the system of algebraic equations for derivatives of unknown coefficients A and B:
\[
\begin{split}
A' \,e^{-ky} + B'\, e^{ky} &= 0 ,
\\
-k\,A' \,e^{-ky} + k\,B'\, e^{ky} &= g(y) .
\end{split}
\]
It is not hard to solve the above system of algebraic equations to obtain
\[
A' = -\frac{1}{2k}\, e^{ky}\, g(y) , \qquad B' = \frac{1}{2k}\, e^{-ky}\, g(y) .
\]
Integrating yields
\[
A(y) = -\frac{1}{2k}\, \int_0^{y} {\text d} \xi \,e^{k\xi}\, g(\xi ) , \qquad B(y) = -\frac{1}{2k}\, \int_y^{\infty} {\text d} \xi \,e^{-k\xi}\, g(\xi ) ,
\]
where we dropped constant of integration that we will incorporate shortly. Using these expressions, we find the general solution that satisfies the regularity condition at infinity to be
\begin{align*}
v(y) &= -\frac{1}{2k}\,\int_0^{\infty} {\text d} \xi \,e^{-k|y - \xi |}\, g(\xi )
+C\, e^{-ky}
\\
&=-\frac{1}{2k}\,e^{-ky} \,\int_0^{y} {\text d} \xi \,e^{k\, \xi}\, g(\xi ) -\frac{1}{2k}\,e^{ky} \,\int_y^{\infty} {\text d} \xi \,e^{-k\, \xi}\, g(\xi ) +C\, e^{-ky} ,
\end{align*}
where C is an arbitrary constant. To satisfy the boundary condition at y = 0, we substitute the above function into this condition to obtain
\[
-\frac{1-k}{2k}\, \int_0^{\infty} {\text d} \xi \,e^{-k\xi}\, g(\xi ) + C \left( 1 - k \right) = 0 .
\]
The above equation can be solved with respect to C to obtain its value and the solution becomes
\[
v(y) = -\frac{1}{2\,k}\,\int_0^{\infty} {\text d} \xi \,e^{-k|y- \xi |}\, g(\xi ) +
\frac{1}{2\,k}\, \int_0^{\infty} {\text d} \xi \,e^{-k \left( y+\xi \right)}\, g(\xi ) .
\]
We cannot apply the inverse cosine-Fourier transform to the above function because it has a pole at k = 0. But we can find its partial derivatives.
\[
\frac{{\text d}v(y)}{{\text d}y} = \frac{1}{2} \, e^{-ky} \int_0^y g(\xi )\, e^{k\xi}\,{\text d}\xi - \frac{1}{2} \, e^{ky} \int_y^{\infty} g(\xi )\, e^{-k\xi}\,{\text d}\xi - \frac{1}{2} \, e^{-ky} \int_y^{\infty} g(\xi )\, e^{-k\xi}\,{\text d}\xi .
\]
Hence
\begin{align*}
\frac{\partial u}{\partial y} &= ℱ_c^{-1} \left[ \frac{{\text d}v(y)}{{\text d}y} \right] = \frac{2}{\pi} \int_0^{\infty} \frac{{\text d}v(y)}{{\text d}y} \,\cos (kx)\,{\text d} k
\\
&= \frac{1}{\pi} \int_0^y {\text d}\xi \,g(\xi )\,\frac{y-\xi}{(y-\xi )^2 + x^2} - \frac{1}{\pi} \int_0^y {\text d}\xi \,g(\xi )\,\frac{\xi - y}{(y-\xi )^2 + x^2} - \frac{1}{\pi} \int_0^{\infty} {\text d}\xi \,g(\xi )\,\frac{y+\xi}{(y+\xi )^2 + x^2}
\\
&= \frac{1}{\pi} \int_0^{\infty} {\text d}\xi \,g(\xi )\,\frac{y-\xi}{(y-\xi )^2 + x^2} - \frac{1}{\pi} \int_0^{\infty} {\text d}\xi \,g(\xi )\,\frac{y+\xi}{(y+\xi )^2 + x^2}.
\end{align*}
Assuming[x > 0 && A > 0,
Integrate[Cos[k*x]*Exp[-A*k], {k, 0, Infinity}]]
A/(A^2 + x^2)
Correspondingly,
\[
\frac{\partial u}{\partial x} = - \frac{1}{\pi} \,\int_0^{\infty} g(\xi )\, \frac{x}{(y-\xi )^2 + x^2} \,{\text d} \xi + \frac{1}{\pi} \,\int_0^{\infty} g(\xi )\, \frac{x}{(y+\xi )^2 + x^2} \,{\text d} \xi .
\]
Integration yields
Assuming[0 < y && s > 0, Integrate[x/((y - s)^2 + x^2), x]]
1/2 Log[s^2 + x^2 - 2 s y + y^2]
Assuming[0 < y && s > 0, Integrate[x/((y + s)^2 + x^2), x]]
1/2 Log[s^2 + x^2 + 2 s y + y^2]
\[
u(x,y) = - \frac{1}{2\pi}\,\int_0^{\infty} g(\xi )\, \ln \left( x^2 + \xi^2 -2x\xi + y^2 \right) {\text d}\xi + \frac{1}{2\pi}\,\int_0^{\infty} g(\xi )\, \ln \left( x^2 + \xi^2 +2x\xi + y^2 \right) {\text d}\xi + \phi (y) ,
\]
where ϕ(y) is unknown function.
Taking derivative with respect to y and equating to known function, we obtain
\[
\frac{\partial u}{\partial y} = \frac{1}{\pi} \,\int_0^{\infty} g(\xi )\, \,\frac{y-\xi}{(y-\xi )^2 + x^2} \,{\text d}\xi - \frac{1}{\pi} \,\int_0^{\infty} g(\xi )\, \,\frac{y+\xi}{(y+\xi )^2 + x^2} \,{\text d}\xi + \phi' (y) .
\]
Assuming[0 < y && s > 0 && x > 0,
D[1/2 Log[s^2 + x^2 - 2 s y + y^2], y]]
(-2 s + 2 y)/(2 (s^2 + x^2 - 2 s y + y^2))
Assuming[0 < y && s > 0 && x > 0,
D[1/2 Log[s^2 + x^2 + 2 s y + y^2], y]]
(2 s + 2 y)/(2 (s^2 + x^2 + 2 s y + y^2))
Therefore, the function ϕ(y) is a constant, which we have to set equal to zero in order to satisfy the boundary conditions. Finally, we have
\[
S_2 \left[ g(y) \right] (x,y) =
- \frac{1}{2\pi}\,\int_0^{\infty} g(\xi )\, \ln \left( x^2 + \xi^2 -2x\xi + y^2 \right) {\text d}\xi + \frac{1}{2\pi}\,\int_0^{\infty} g(\xi )\, \ln \left( x^2 + \xi^2 +2x\xi + y^2 \right) {\text d}\xi .
\]
Its projection on the boundary x = 0 is
\[
B_1 S_2 \left[ g(y) \right] (x,y) = \lim_{x\to 0} S_2 \left[ g(y) \right] (x,y) = 0.
\]
The resolvent method solves the required boundary value problem
We are looking for solution of the given boundary value problem with mixed boundary conditions in the form
\[
u(x,y) = \left( S_1 + S_2 \right) \varphi ,
\]
where unknown function φ(y) should be determined from the equation
\[
\left( B_1 - B_2 \right) \left( S_1 + S_2 \right) \varphi = g ,
\]
or
\[
\frac{1}{\pi}\,\int_{0}^{\infty} \frac{\varphi (\xi ) - \varphi (y)}{\left( \xi - y \right)^2} \,{\text d} \xi - \frac{1}{\pi}\,\int_{0}^{\infty} \frac{\varphi (\xi ) - \varphi (0)}{\left( \xi + y \right)^2} \,{\text d} \xi + \frac{\varphi (0)}{\pi\,y} = -g(y) .
\]
Assuming that g(0) = φ(0) = 0, we can drop the term φ(0) in the above integral equation.
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