The Poisson equation, \( -\Delta u = f \), is one of the most fundamental elliptic partial differential equations in mathematical physics and analysis. Mastering it requires working through a range of problems: from direct verification of classical solutions and Dirichlet boundary value problems, to Green’s functions, fundamental solutions, superposition principles, Neumann conditions, and weak (variational) formulations. The exercises below are organized by theme and progress from straightforward computations to graduate-level analysis, covering key LSI topics such as harmonic functions, the Laplace operator, energy estimates, uniqueness, and the method of images.
Verification and Direct Solutions of the Poisson Equation
These problems build the essential skill of verifying whether a given function satisfies \(-\Delta u = f\) on a domain, computing the Laplacian in Cartesian and polar coordinates, and identifying source terms. Mastering direct verification is the indispensable first step before tackling boundary value problems or Green’s function theory.
Problem 1: Verifying a Polynomial Solution
Easy
Consider the function \( u(x,y) = x^2 – y^2 \) defined on \(\mathbb{R}^2\).
- Compute \(\Delta u\) and determine whether \(u\) satisfies the Poisson equation \(-\Delta u = f\) for some \(f\). Identify \(f\).
- Now consider \(v(x,y) = x^2 + y^2\). Compute \(-\Delta v\) and state what equation \(v\) satisfies.
Hint
View Solution
\[ \partial_{xx}(x^2 – y^2) = 2, \quad \partial_{yy}(x^2 – y^2) = -2 \]
\[ \Delta u = 2 + (-2) = 0 \]
So \(-\Delta u = 0\), meaning \(u\) is harmonic; it satisfies the Laplace equation, which is the special case of the Poisson equation with \(f = 0\).
Solution to question 2:
\[ \partial_{xx}(x^2 + y^2) = 2, \quad \partial_{yy}(x^2 + y^2) = 2 \]
\[ \Delta v = 4, \quad -\Delta v = -4 \]
Thus \(v\) satisfies the Poisson equation \(-\Delta v = -4\), i.e., with source term \(f \equiv -4\).
Problem 2: Verification in Polar Coordinates
Easy
Let \( u(r) = r^2 \ln r \) for \( r > 0 \), where \( r = \sqrt{x^2 + y^2} \) is the radial variable in \(\mathbb{R}^2\).
- Using the radial Laplacian \(\Delta u = u”(r) + \frac{1}{r}u'(r)\), compute \(\Delta u\).
- Write the corresponding Poisson equation and identify the source term \(f(r)\).
Hint
View Solution
\[ u'(r) = 2r\ln r + r, \quad u”(r) = 2\ln r + 3 \]
\[ \Delta u = (2\ln r + 3) + \frac{1}{r}(2r\ln r + r) = 2\ln r + 3 + 2\ln r + 1 = 4\ln r + 4 \]
Solution to question 2:
\[ -\Delta u = -(4\ln r + 4) \]
The Poisson equation satisfied by \(u\) is \(-\Delta u = f(r)\) with \(f(r) = -4(\ln r + 1)\).
Problem 3: Constructing a Source Term from a Given Solution
Medium
Suppose you want \( u(x,y,z) = e^{x}\sin(y)\cos(z) \) to be the exact solution of a Poisson equation \(-\Delta u = f\) on \(\mathbb{R}^3\).
- Compute each second-order partial derivative of \(u\) and find \(\Delta u\).
- Determine the explicit source term \(f(x,y,z)\) such that \(u\) is the solution.
Hint
View Solution
\[ \partial_{xx} u = e^x \sin y \cos z \]
\[ \partial_{yy} u = -e^x \sin y \cos z \]
\[ \partial_{zz} u = -e^x \sin y \cos z \]
\[ \Delta u = e^x \sin y \cos z – e^x \sin y \cos z – e^x \sin y \cos z = -e^x \sin y \cos z \]
Solution to question 2:
\[ -\Delta u = e^x \sin y \cos z \]
Therefore \( f(x,y,z) = e^x \sin y \cos z \), and \(u\) itself equals \(f\). This is a self-referential Poisson equation where the source equals the solution.
Dirichlet Boundary Value Problems for the Poisson Equation
The Dirichlet problem asks for a function \(u\) satisfying \(-\Delta u = f\) in a domain \(\Omega\) with prescribed values \(u = g\) on the boundary \(\partial\Omega\). These exercises develop the core techniques: separation of variables, eigenfunction expansion, and verification of uniqueness via energy methods — all central to PDE courses at undergraduate and graduate level.
Problem 4: Poisson Equation on an Interval
Easy
Consider the one-dimensional Poisson equation on \((0,1)\):
\[ -u”(x) = f(x), \quad x \in (0,1), \quad u(0) = 0,\; u(1) = 0 \]
with \(f(x) = 2\).
- Find the general solution of \(-u” = 2\) and apply the boundary conditions to find the unique solution.
- Verify your answer by substituting back into the equation and checking the boundary conditions.
Hint
View Solution
Integrating twice: \(u(x) = -x^2 + Ax + B\).
\[ u(0) = 0 \Rightarrow B = 0 \]
\[ u(1) = -1 + A = 0 \Rightarrow A = 1 \]
\[ \boxed{u(x) = x(1-x)} \]
Solution to question 2:
\(-u”(x) = -(-2) = 2 = f(x)\;\checkmark\)
\(u(0) = 0\;\checkmark,\quad u(1) = 1(1-1) = 0\;\checkmark\)
Problem 5: Uniqueness via the Energy Method
Medium
Let \(\Omega \subset \mathbb{R}^n\) be a bounded domain with smooth boundary. Suppose \(u_1\) and \(u_2\) are two classical solutions of:
\[ -\Delta u = f \text{ in } \Omega, \quad u = g \text{ on } \partial\Omega \]
- Define \(w = u_1 – u_2\). Write the boundary value problem satisfied by \(w\).
- Multiply the equation for \(w\) by \(w\) itself, integrate over \(\Omega\), apply Green’s first identity, and conclude that \(w \equiv 0\), proving uniqueness.
Hint
View Solution
Since both \(u_1\) and \(u_2\) solve the same equation with the same data:
\[ -\Delta w = 0 \text{ in } \Omega, \quad w = 0 \text{ on } \partial\Omega \]
Solution to question 2:
Multiply \(-\Delta w = 0\) by \(w\) and integrate:
\[ \int_\Omega w(-\Delta w)\,dx = 0 \]
By Green’s first identity and \(w|_{\partial\Omega}=0\):
\[ \int_\Omega |\nabla w|^2\,dx – \underbrace{\int_{\partial\Omega} w\,\partial_\nu w\,dS}_{=\,0} = 0 \]
\[ \Rightarrow \int_\Omega |\nabla w|^2\,dx = 0 \Rightarrow \nabla w = 0 \text{ a.e.} \Rightarrow w \equiv C \]
Since \(w=0\) on \(\partial\Omega\) and \(\Omega\) is connected, \(C=0\), so \(w \equiv 0\) and \(u_1 = u_2\).
Green’s Functions and Fundamental Solutions
The fundamental solution of the Laplacian and domain-specific Green’s functions provide explicit solution formulas for the Poisson equation. These problems train students to derive the fundamental solution in \(\mathbb{R}^n\), construct Green’s functions for simple domains via the method of images, and apply the Poisson integral formula — skills essential in both theoretical PDE and applied mathematics.
Problem 6: The Fundamental Solution in \(\mathbb{R}^3\)
Medium
The fundamental solution of \(-\Delta\) in \(\mathbb{R}^3\) is \(\Phi(x) = \dfrac{1}{4\pi |x|}\).
- Show that \(\Phi\) is harmonic (i.e., \(\Delta\Phi = 0\)) for \(x \neq 0\) by direct computation in spherical coordinates.
- Given a continuous, compactly supported function \(f\), write the convolution formula \(u = \Phi * f\) and state what Poisson equation \(u\) satisfies.
Hint
View Solution
With \(\Phi(r) = \frac{1}{4\pi r}\):
\[ \Phi'(r) = -\frac{1}{4\pi r^2}, \quad \Phi”(r) = \frac{2}{4\pi r^3} = \frac{1}{2\pi r^3} \]
\[ \Delta\Phi = \frac{1}{2\pi r^3} + \frac{2}{r}\cdot\left(-\frac{1}{4\pi r^2}\right) = \frac{1}{2\pi r^3} – \frac{1}{2\pi r^3} = 0 \quad (r\neq 0) \]
Solution to question 2:
\[ u(x) = (\Phi * f)(x) = \int_{\mathbb{R}^3} \frac{f(y)}{4\pi|x-y|}\,dy \]
This function satisfies \(-\Delta u = f\) in \(\mathbb{R}^3\) (in the distributional sense, using \(-\Delta\Phi = \delta_0\)).
Problem 7: Green’s Function for the Half-Space via the Method of Images
Medium
Let \(\Omega = \{x = (x_1,x_2,x_3)\in\mathbb{R}^3 : x_3 > 0\}\) be the upper half-space. For a source point \(y = (y_1,y_2,y_3)\) with \(y_3 > 0\), define the reflected point \(\tilde{y} = (y_1,y_2,-y_3)\).
- Write the Green’s function \(G(x,y)\) for the Dirichlet Laplacian on \(\Omega\) using the method of images, and verify that \(G(x,y) = 0\) for \(x_3 = 0\).
- For \(f \equiv 0\), write the Poisson integral formula giving the solution \(u\) in terms of boundary data \(g(x_1,x_2) = u(x_1,x_2,0)\).
Hint
View Solution
\[ G(x,y) = \frac{1}{4\pi|x-y|} – \frac{1}{4\pi|x-\tilde{y}|} \]
On \(x_3 = 0\): \(|x – y|^2 = (x_1-y_1)^2+(x_2-y_2)^2+y_3^2\) and \(|x-\tilde{y}|^2 = (x_1-y_1)^2+(x_2-y_2)^2+y_3^2\), so \(|x-y|=|x-\tilde{y}|\) and \(G = 0\;\checkmark\)
Solution to question 2:
The Poisson integral formula for the half-space is:
\[ u(x) = \frac{x_3}{2\pi}\int_{\mathbb{R}^2} \frac{g(y_1,y_2)}{\left[(x_1-y_1)^2+(x_2-y_2)^2+x_3^2\right]^{3/2}}\,dy_1\,dy_2 \]
This is obtained by differentiating \(G\) with respect to the outward normal on \(\partial\Omega\) (the Poisson kernel).
Problem 8: Green’s Function for the Unit Ball in \(\mathbb{R}^3\)
Hard
Let \(B = B(0,1) \subset \mathbb{R}^3\) be the open unit ball. For \(y \in B\), define the Kelvin inversion \(\tilde{y} = y/|y|^2\).
- Write the Green’s function for the Dirichlet Laplacian on \(B\) and verify the boundary condition \(G(x,y)=0\) for \(|x|=1\).
- Using the outward normal derivative of \(G\), derive the Poisson kernel \(K(x,y)\) for the unit ball and write the integral formula for the harmonic extension of boundary data \(g \in C(\partial B)\).
Hint
View Solution
\[ G(x,y) = \frac{1}{4\pi|x-y|} – \frac{1}{4\pi\,|y|\,\left|x – \tilde{y}\right|} \]
For \(|x|=1\): one verifies that \(|x-\tilde{y}| = |x-y|/|y|\), so:
\[ G(x,y) = \frac{1}{4\pi|x-y|} – \frac{|y|}{4\pi\,|y|\,|x-y|} = 0\;\checkmark \]
Solution to question 2:
The Poisson kernel (outward normal derivative on \(\partial B\)) is:
\[ K(x,y) = -\partial_{\nu_x} G(x,y)\big|_{|x|=1} = \frac{1-|y|^2}{4\pi|x-y|^3} \]
The harmonic extension of \(g\) is:
\[ u(y) = \int_{\partial B} K(x,y)\,g(x)\,dS_x = \frac{1-|y|^2}{4\pi}\int_{\partial B} \frac{g(x)}{|x-y|^3}\,dS_x \]
Neumann Boundary Conditions and Mixed Problems
While Dirichlet conditions fix the values of the solution on \(\partial\Omega\), Neumann boundary conditions prescribe the normal derivative \(\partial_\nu u\). These problems expose a critical compatibility condition for existence, explore uniqueness up to a constant, and introduce mixed (Robin) boundary value problems — all of which appear in heat conduction, electrostatics, and fluid mechanics.
Problem 9: Compatibility Condition for the Neumann Problem
Medium
Let \(\Omega\subset\mathbb{R}^n\) be a bounded domain. Consider the Neumann problem:
\[ -\Delta u = f \text{ in } \Omega, \quad \partial_\nu u = h \text{ on } \partial\Omega \]
- By integrating the equation over \(\Omega\) and applying the Divergence Theorem, derive the necessary compatibility condition relating \(f\) and \(h\).
- Show that if \(u\) is a solution, then \(u + C\) is also a solution for any constant \(C\), and state what additional condition makes the solution unique.
Hint
View Solution
\[ \int_\Omega f\,dx = -\int_\Omega \Delta u\,dx = -\int_{\partial\Omega} \partial_\nu u\,dS = -\int_{\partial\Omega} h\,dS \]
The necessary compatibility condition is:
\[ \boxed{\int_\Omega f\,dx + \int_{\partial\Omega} h\,dS = 0} \]
Solution to question 2:
If \(-\Delta u = f\), then \(-\Delta(u+C) = -\Delta u = f\) and \(\partial_\nu(u+C) = \partial_\nu u = h\), so \(u+C\) is also a solution. To ensure uniqueness, impose a normalization, e.g., \(\displaystyle\int_\Omega u\,dx = 0\), which fixes the additive constant.
Problem 10: Mixed Dirichlet–Neumann Problem on a Rectangle
Hard
On the rectangle \(\Omega=(0,L)\times(0,H)\), solve the following mixed problem using separation of variables and superposition:
\[ -\Delta u = 0 \text{ in } \Omega \]
with boundary conditions: \(u(0,y) = 0\), \(u(L,y) = 0\), \(u(x,0) = 0\), and \(\partial_y u(x,H) = g(x)\).
- Seek separated solutions and derive the eigenfunctions satisfying the three homogeneous conditions.
- Expand \(g(x)\) in the appropriate eigenfunction basis and write the formal series solution \(u(x,y)\).
Hint
View Solution
Separating \(-\Delta u = 0\) with \(u=X(x)Y(y)\):
\[ X” + \lambda X = 0,\quad X(0)=X(L)=0 \Rightarrow X_n(x)=\sin\!\left(\tfrac{n\pi x}{L}\right),\;\lambda_n = \left(\tfrac{n\pi}{L}\right)^2 \]
\[ Y” – \lambda_n Y = 0,\quad Y_n(0)=0 \Rightarrow Y_n(y) = \sinh\!\left(\tfrac{n\pi y}{L}\right) \]
Solution to question 2:
\[ u(x,y) = \sum_{n=1}^\infty A_n \sin\!\left(\tfrac{n\pi x}{L}\right)\sinh\!\left(\tfrac{n\pi y}{L}\right) \]
The Neumann condition \(\partial_y u(x,H) = g(x)\) gives:
\[ \sum_{n=1}^\infty A_n \frac{n\pi}{L}\cosh\!\left(\tfrac{n\pi H}{L}\right)\sin\!\left(\tfrac{n\pi x}{L}\right) = g(x) \]
Multiplying by \(\sin(m\pi x/L)\) and integrating over \((0,L)\):
\[ A_n = \frac{2}{n\pi\cosh(n\pi H/L)}\int_0^L g(x)\sin\!\left(\tfrac{n\pi x}{L}\right)dx \]
Weak Formulation and Variational Methods
The weak (variational) formulation of the Poisson equation is the gateway to modern PDE analysis and numerical methods such as the finite element method. These problems guide students from multiplying by a test function and integrating by parts, to establishing well-posedness via the Lax–Milgram theorem and the Poincaré–Friedrichs inequality — topics at the heart of functional analysis and computational mathematics.
Problem 11: Deriving the Weak Formulation
Medium
Let \(\Omega\subset\mathbb{R}^n\) be a bounded domain with smooth boundary, and consider:
\[ -\Delta u = f \text{ in } \Omega, \quad u = 0 \text{ on } \partial\Omega \]
with \(f\in L^2(\Omega)\).
- Multiply the equation by a test function \(v \in H^1_0(\Omega)\), integrate over \(\Omega\), and apply Green’s first identity to derive the weak formulation: find \(u\in H^1_0(\Omega)\) such that \(a(u,v) = F(v)\) for all \(v\in H^1_0(\Omega)\). Identify the bilinear form \(a(\cdot,\cdot)\) and the linear functional \(F\).
- State the Lax–Milgram theorem conditions that guarantee a unique weak solution, and verify them for this bilinear form using the Poincaré inequality.
Hint
View Solution
Multiply \(-\Delta u = f\) by \(v\in H^1_0(\Omega)\) and integrate:
\[ \int_\Omega (-\Delta u)v\,dx = \int_\Omega fv\,dx \]
By Green’s first identity (with \(v=0\) on \(\partial\Omega\)):
\[ \int_\Omega \nabla u\cdot\nabla v\,dx = \int_\Omega fv\,dx \]
The weak formulation is: find \(u\in H^1_0(\Omega)\) such that
\[ a(u,v) = F(v)\quad\forall\,v\in H^1_0(\Omega) \]
with \(a(u,v) = \displaystyle\int_\Omega\nabla u\cdot\nabla v\,dx\) and \(F(v)=\displaystyle\int_\Omega fv\,dx\).
Solution to question 2:
Lax–Milgram requires: (i) continuity: \(|a(u,v)|\leq\|\nabla u\|_{L^2}\|\nabla v\|_{L^2}\leq\|u\|_{H^1}\|v\|_{H^1}\) ✓; (ii) coercivity: \(a(u,u)=\|\nabla u\|^2_{L^2}\geq C\|u\|^2_{H^1}\) by the Poincaré–Friedrichs inequality \(\|u\|_{L^2}\leq C_P\|\nabla u\|_{L^2}\) for \(u\in H^1_0(\Omega)\) ✓. Both conditions hold, so a unique weak solution exists.
Problem 12: Inhomogeneous Dirichlet Data in the Weak Setting
Hard
Consider the Poisson equation with nonhomogeneous Dirichlet data:
\[ -\Delta u = f \text{ in } \Omega, \quad u = g \text{ on } \partial\Omega \]
where \(f\in L^2(\Omega)\) and \(g\in H^1(\Omega)\) (understood as a trace).
- Introduce the lifting \(\tilde{g}\in H^1(\Omega)\) with \(\tilde{g}|_{\partial\Omega} = g\), define \(w = u – \tilde{g}\), and write the equation satisfied by \(w\) with homogeneous Dirichlet data.
- Write the weak formulation for \(w\) and invoke the Lax–Milgram theorem to conclude existence and uniqueness of the weak solution \(u\).
Hint
View Solution
Set \(w = u – \tilde{g}\). Then \(-\Delta w = -\Delta u + \Delta\tilde{g} = f + \Delta\tilde{g}\) in \(\Omega\), and \(w = u – \tilde{g} = g – g = 0\) on \(\partial\Omega\). The new source is \(\tilde{f} = f + \Delta\tilde{g}\in H^{-1}(\Omega)\).
Solution to question 2:
The weak formulation for \(w\in H^1_0(\Omega)\) is:
\[ \int_\Omega \nabla w\cdot\nabla v\,dx = \int_\Omega fv\,dx – \int_\Omega \nabla\tilde{g}\cdot\nabla v\,dx \quad \forall\,v\in H^1_0(\Omega) \]
The right-hand side defines a bounded linear functional on \(H^1_0(\Omega)\) (by Cauchy–Schwarz). Since \(a(w,v)=\int_\Omega\nabla w\cdot\nabla v\,dx\) is continuous and coercive (as shown in Problem 12), Lax–Milgram guarantees a unique \(w\in H^1_0(\Omega)\), and thus a unique weak solution \(u = w + \tilde{g}\in H^1(\Omega)\).
Superposition, Maximum Principle, and Applications
The superposition principle and the maximum principle are two of the most powerful tools for understanding the Poisson and Laplace equations.
Problem 13: Maximum Principle and Comparison
Hard
Let \(\Omega\subset\mathbb{R}^n\) be a bounded connected domain, and let \(u\in C^2(\Omega)\cap C(\overline{\Omega})\) satisfy \(-\Delta u \leq 0\) in \(\Omega\) (i.e., \(u\) is subharmonic).
- State the weak maximum principle for subharmonic functions and prove it by contradiction: assume the maximum is achieved at an interior point \(x_0\) and derive a contradiction using the second-order condition \(\Delta u(x_0) \geq 0\).
- Use the maximum principle to prove: if \(u_1, u_2\in C^2(\Omega)\cap C(\overline{\Omega})\) both solve \(-\Delta u = f\) in \(\Omega\) with \(u_1 \leq u_2\) on \(\partial\Omega\), then \(u_1\leq u_2\) throughout \(\overline{\Omega}\).
Hint
View Solution
Weak Maximum Principle: \(\max_{\overline{\Omega}} u = \max_{\partial\Omega} u\).
Proof: Suppose for contradiction that \(u\) attains its global maximum at an interior point \(x_0\in\Omega\). Then by the second-order necessary condition for a maximum: \(\partial_{x_i x_i}u(x_0)\leq 0\) for each \(i\), so \(\Delta u(x_0)\leq 0\). But \(-\Delta u\leq 0\) means \(\Delta u(x_0)\geq 0\). Hence \(\Delta u(x_0) = 0\). A more careful argument (using the mean value inequality for subharmonic functions) shows \(u\) must be constant in a neighborhood of \(x_0\), and by connectedness, \(u \equiv \text{const}\) on \(\Omega\), contradicting a strict interior maximum. Thus \(\max_{\overline{\Omega}}u = \max_{\partial\Omega}u\).
Solution to question 2:
Let \(w = u_2 – u_1\). Then \(-\Delta w = -\Delta u_2 + \Delta u_1 = f – f = 0\) in \(\Omega\), so \(w\) is harmonic, hence both subharmonic and superharmonic. By the maximum principle applied to \(w\):
\[ \max_{\overline{\Omega}} w = \max_{\partial\Omega} w \leq 0 \]
since \(u_1 \leq u_2\) on \(\partial\Omega\) gives \(w \leq 0\) on \(\partial\Omega\). Therefore \(w = u_2 – u_1 \leq 0\) on \(\overline{\Omega}\), i.e., \(u_1\leq u_2\) throughout \(\overline{\Omega}\).