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Poisson Equation: Problems & Exercises with Solutions

Type
Practice problems
Problems
13
Levels
Easy · Medium · Hard

The Poisson equation, −Δu=f -\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.

Contents6 sections

Verification and Direct Solutions of the Poisson Equation

These problems build the essential skill of verifying whether a given function satisfies −Δu=f-\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)=x2−y2 u(x,y) = x^2 - y^2 defined on R2\mathbb{R}^2.

  1. Compute Δu\Delta u and determine whether uu satisfies the Poisson equation −Δu=f-\Delta u = f for some ff. Identify ff.
  2. Now consider v(x,y)=x2+y2v(x,y) = x^2 + y^2. Compute −Δv-\Delta v and state what equation vv satisfies.
Hint
Recall that for u(x,y)u(x,y), the Laplacian is Δu=∂xxu+∂yyu\Delta u = \partial_{xx} u + \partial_{yy} u. Compute each second partial derivative separately.
View Solution
Solution to question 1: ∂xx(x2−y2)=2,∂yy(x2−y2)=−2 \partial_{xx}(x^2 - y^2) = 2, \quad \partial_{yy}(x^2 - y^2) = -2 Δu=2+(−2)=0 \Delta u = 2 + (-2) = 0 So −Δu=0-\Delta u = 0, meaning uu is harmonic; it satisfies the Laplace equation, which is the special case of the Poisson equation with f=0f = 0. Solution to question 2: ∂xx(x2+y2)=2,∂yy(x2+y2)=2 \partial_{xx}(x^2 + y^2) = 2, \quad \partial_{yy}(x^2 + y^2) = 2 Δv=4,−Δv=−4 \Delta v = 4, \quad -\Delta v = -4 Thus vv satisfies the Poisson equation −Δv=−4-\Delta v = -4, i.e., with source term f≡−4f \equiv -4.

Problem 2: Verification in Polar Coordinates

Easy

Let u(r)=r2ln⁡r u(r) = r^2 \ln r for r>0 r > 0 , where r=x2+y2 r = \sqrt{x^2 + y^2} is the radial variable in R2\mathbb{R}^2.

  1. Using the radial Laplacian Δu=u′′(r)+1ru′(r)\Delta u = u''(r) + \frac{1}{r}u'(r), compute Δu\Delta u.
  2. Write the corresponding Poisson equation and identify the source term f(r)f(r).
Hint
Compute u′(r)u'(r) and u′′(r)u''(r) carefully using the product rule. Substitute into the radial form of the Laplacian.
View Solution
Solution to question 1: u′(r)=2rln⁡r+r,u′′(r)=2ln⁡r+3 u'(r) = 2r\ln r + r, \quad u''(r) = 2\ln r + 3 Δu=(2ln⁡r+3)+1r(2rln⁡r+r)=2ln⁡r+3+2ln⁡r+1=4ln⁡r+4 \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: −Δu=−(4ln⁡r+4) -\Delta u = -(4\ln r + 4) The Poisson equation satisfied by uu is −Δu=f(r)-\Delta u = f(r) with f(r)=−4(ln⁡r+1)f(r) = -4(\ln r + 1).

Problem 3: Constructing a Source Term from a Given Solution

Medium

Suppose you want u(x,y,z)=exsin⁡(y)cos⁡(z) u(x,y,z) = e^{x}\sin(y)\cos(z) to be the exact solution of a Poisson equation −Δu=f-\Delta u = f on R3\mathbb{R}^3.

  1. Compute each second-order partial derivative of uu and find Δu\Delta u.
  2. Determine the explicit source term f(x,y,z)f(x,y,z) such that uu is the solution.
Hint
Differentiate uu twice with respect to xx, yy, and zz separately, then sum the results. Note that ∂xxex=ex\partial_{xx}e^x = e^x, ∂yysin⁡y=−sin⁡y\partial_{yy}\sin y = -\sin y, and ∂zzcos⁡z=−cos⁡z\partial_{zz}\cos z = -\cos z.
View Solution
Solution to question 1: ∂xxu=exsin⁡ycos⁡z \partial_{xx} u = e^x \sin y \cos z ∂yyu=−exsin⁡ycos⁡z \partial_{yy} u = -e^x \sin y \cos z ∂zzu=−exsin⁡ycos⁡z \partial_{zz} u = -e^x \sin y \cos z Δu=exsin⁡ycos⁡z−exsin⁡ycos⁡z−exsin⁡ycos⁡z=−exsin⁡ycos⁡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: −Δu=exsin⁡ycos⁡z -\Delta u = e^x \sin y \cos z Therefore f(x,y,z)=exsin⁡ycos⁡z f(x,y,z) = e^x \sin y \cos z , and uu itself equals ff. 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 uu satisfying −Δu=f-\Delta u = f in a domain Ω\Omega with prescribed values u=gu = g on the boundary ∂Ω\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)(0,1):

−u′′(x)=f(x),x∈(0,1),u(0)=0,  u(1)=0 -u''(x) = f(x), \quad x \in (0,1), \quad u(0) = 0,\; u(1) = 0

with f(x)=2f(x) = 2.

  1. Find the general solution of −u′′=2-u'' = 2 and apply the boundary conditions to find the unique solution.
  2. Verify your answer by substituting back into the equation and checking the boundary conditions.
Hint
Integrate −u′′=2-u'' = 2 twice to get the general solution u(x)=−x2+Ax+Bu(x) = -x^2 + Ax + B. Then use the two boundary conditions to determine the constants AA and BB.
View Solution
Solution to question 1: Integrating twice: u(x)=−x2+Ax+Bu(x) = -x^2 + Ax + B. u(0)=0⇒B=0 u(0) = 0 \Rightarrow B = 0 u(1)=−1+A=0⇒A=1 u(1) = -1 + A = 0 \Rightarrow A = 1 u(x)=x(1−x) \boxed{u(x) = x(1-x)} Solution to question 2: −u′′(x)=−(−2)=2=f(x)  ✓-u''(x) = -(-2) = 2 = f(x)\;\checkmark u(0)=0  ✓,u(1)=1(1−1)=0  ✓u(0) = 0\;\checkmark,\quad u(1) = 1(1-1) = 0\;\checkmark

Problem 5: Uniqueness via the Energy Method

Medium

Let Ω⊂Rn\Omega \subset \mathbb{R}^n be a bounded domain with smooth boundary. Suppose u1u_1 and u2u_2 are two classical solutions of: −Δu=f in Ω,u=g on ∂Ω -\Delta u = f \text{ in } \Omega, \quad u = g \text{ on } \partial\Omega

  1. Define w=u1−u2w = u_1 - u_2. Write the boundary value problem satisfied by ww.
  2. Multiply the equation for ww by ww itself, integrate over Ω\Omega, apply Green’s first identity, and conclude that w≡0w \equiv 0, proving uniqueness.
Hint
After subtracting the two equations, ww satisfies −Δw=0-\Delta w = 0 with w=0w = 0 on ∂Ω\partial\Omega. Use Green’s first identity: ∫Ωw(−Δw) dx=∫Ω∣∇w∣2 dx−∫∂Ωw ∂νw dS\int_\Omega w(-\Delta w)\,dx = \int_\Omega |\nabla w|^2\,dx - \int_{\partial\Omega} w\,\partial_\nu w\,dS.
View Solution
Solution to question 1: Since both u1u_1 and u2u_2 solve the same equation with the same data: −Δw=0 in Ω,w=0 on ∂Ω -\Delta w = 0 \text{ in } \Omega, \quad w = 0 \text{ on } \partial\Omega Solution to question 2: Multiply −Δw=0-\Delta w = 0 by ww and integrate: ∫Ωw(−Δw) dx=0 \int_\Omega w(-\Delta w)\,dx = 0 By Green’s first identity and w∣∂Ω=0w|_{\partial\Omega}=0: ∫Ω∣∇w∣2 dx−∫∂Ωw ∂νw dS⏟= 0=0 \int_\Omega |\nabla w|^2\,dx - \underbrace{\int_{\partial\Omega} w\,\partial_\nu w\,dS}_{=\,0} = 0 ⇒∫Ω∣∇w∣2 dx=0⇒∇w=0 a.e.⇒w≡C \Rightarrow \int_\Omega |\nabla w|^2\,dx = 0 \Rightarrow \nabla w = 0 \text{ a.e.} \Rightarrow w \equiv C Since w=0w=0 on ∂Ω\partial\Omega and Ω\Omega is connected, C=0C=0, so w≡0w \equiv 0 and u1=u2u_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 Rn\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 R3\mathbb{R}^3

Medium

The fundamental solution of −Δ-\Delta in R3\mathbb{R}^3 is Φ(x)=14π∣x∣\Phi(x) = \dfrac{1}{4\pi |x|}.

  1. Show that Φ\Phi is harmonic (i.e., ΔΦ=0\Delta\Phi = 0) for x≠0x \neq 0 by direct computation in spherical coordinates.
  2. Given a continuous, compactly supported function ff, write the convolution formula u=Φ∗fu = \Phi * f and state what Poisson equation uu satisfies.
Hint
In R3\mathbb{R}^3, for a radial function Φ(r)\Phi(r), the Laplacian is ΔΦ=Φ′′(r)+2rΦ′(r)\Delta\Phi = \Phi''(r) + \frac{2}{r}\Phi'(r). Compute Φ′(r)\Phi'(r) and Φ′′(r)\Phi''(r) for Φ(r)=14πr\Phi(r) = \frac{1}{4\pi r}.
View Solution
Solution to question 1: With Φ(r)=14πr\Phi(r) = \frac{1}{4\pi r}: Φ′(r)=−14πr2,Φ′′(r)=24πr3=12πr3 \Phi'(r) = -\frac{1}{4\pi r^2}, \quad \Phi''(r) = \frac{2}{4\pi r^3} = \frac{1}{2\pi r^3} ΔΦ=12πr3+2r⋅(−14πr2)=12πr3−12πr3=0(r≠0) \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)=(Φ∗f)(x)=∫R3f(y)4π∣x−y∣ dy u(x) = (\Phi * f)(x) = \int_{\mathbb{R}^3} \frac{f(y)}{4\pi|x-y|}\,dy This function satisfies −Δu=f-\Delta u = f in R3\mathbb{R}^3 (in the distributional sense, using −ΔΦ=δ0-\Delta\Phi = \delta_0).

Problem 7: Green’s Function for the Half-Space via the Method of Images

Medium

Let Ω={x=(x1,x2,x3)∈R3:x3>0}\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=(y1,y2,y3)y = (y_1,y_2,y_3) with y3>0y_3 > 0, define the reflected point y~=(y1,y2,−y3)\tilde{y} = (y_1,y_2,-y_3).

  1. Write the Green’s function G(x,y)G(x,y) for the Dirichlet Laplacian on Ω\Omega using the method of images, and verify that G(x,y)=0G(x,y) = 0 for x3=0x_3 = 0.
  2. For f≡0f \equiv 0, write the Poisson integral formula giving the solution uu in terms of boundary data g(x1,x2)=u(x1,x2,0)g(x_1,x_2) = u(x_1,x_2,0).
Hint
Place an image charge of opposite sign at the reflected point y~\tilde{y}. The Green’s function is G(x,y)=Φ(x−y)−Φ(x−y~)G(x,y) = \Phi(x-y) - \Phi(x-\tilde{y}). On the boundary x3=0x_3=0, check that ∣x−y∣=∣x−y~∣|x-y|=|x-\tilde{y}|.
View Solution
Solution to question 1: G(x,y)=14π∣x−y∣−14π∣x−y~∣ G(x,y) = \frac{1}{4\pi|x-y|} - \frac{1}{4\pi|x-\tilde{y}|} On x3=0x_3 = 0: ∣x−y∣2=(x1−y1)2+(x2−y2)2+y32|x - y|^2 = (x_1-y_1)^2+(x_2-y_2)^2+y_3^2 and ∣x−y~∣2=(x1−y1)2+(x2−y2)2+y32|x-\tilde{y}|^2 = (x_1-y_1)^2+(x_2-y_2)^2+y_3^2, so ∣x−y∣=∣x−y~∣|x-y|=|x-\tilde{y}| and G=0  ✓G = 0\;\checkmark Solution to question 2: The Poisson integral formula for the half-space is: u(x)=x32π∫R2g(y1,y2)[(x1−y1)2+(x2−y2)2+x32]3/2 dy1 dy2 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 GG with respect to the outward normal on ∂Ω\partial\Omega (the Poisson kernel).

Problem 8: Green’s Function for the Unit Ball in R3\mathbb{R}^3

Hard

Let B=B(0,1)⊂R3B = B(0,1) \subset \mathbb{R}^3 be the open unit ball. For y∈By \in B, define the Kelvin inversion y~=y/∣y∣2\tilde{y} = y/|y|^2.

  1. Write the Green’s function for the Dirichlet Laplacian on BB and verify the boundary condition G(x,y)=0G(x,y)=0 for ∣x∣=1|x|=1.
  2. Using the outward normal derivative of GG, derive the Poisson kernel K(x,y)K(x,y) for the unit ball and write the integral formula for the harmonic extension of boundary data g∈C(∂B)g \in C(\partial B).
Hint
The Green’s function is G(x,y)=Φ(x−y)−∣y∣ Φ(∣y∣x−y/∣y∣)G(x,y) = \Phi(x-y) - |y|\,\Phi(|y|x - y/|y|). On ∣x∣=1|x|=1, use the identity ∣∣y∣x−y/∣y∣ ∣2=∣y∣2∣x−y∣2/∣y∣2||y|x - y/|y|\,|^2 = |y|^2|x-y|^2/|y|^2 (after simplification) to show cancellation. Then compute −∂νG∣∂B-\partial_\nu G|_{\partial B}.
View Solution
Solution to question 1: G(x,y)=14π∣x−y∣−14π ∣y∣ ∣x−y~∣ G(x,y) = \frac{1}{4\pi|x-y|} - \frac{1}{4\pi\,|y|\,\left|x - \tilde{y}\right|} For ∣x∣=1|x|=1: one verifies that ∣x−y~∣=∣x−y∣/∣y∣|x-\tilde{y}| = |x-y|/|y|, so: G(x,y)=14π∣x−y∣−∣y∣4π ∣y∣ ∣x−y∣=0  ✓ 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 ∂B\partial B) is: K(x,y)=−∂νxG(x,y)∣∣x∣=1=1−∣y∣24π∣x−y∣3 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 gg is: u(y)=∫∂BK(x,y) g(x) dSx=1−∣y∣24π∫∂Bg(x)∣x−y∣3 dSx 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 ∂Ω\partial\Omega, Neumann boundary conditions prescribe the normal derivative ∂νu\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 Ω⊂Rn\Omega\subset\mathbb{R}^n be a bounded domain. Consider the Neumann problem: −Δu=f in Ω,∂νu=h on ∂Ω -\Delta u = f \text{ in } \Omega, \quad \partial_\nu u = h \text{ on } \partial\Omega

  1. By integrating the equation over Ω\Omega and applying the Divergence Theorem, derive the necessary compatibility condition relating ff and hh.
  2. Show that if uu is a solution, then u+Cu + C is also a solution for any constant CC, and state what additional condition makes the solution unique.
Hint
Integrate −Δu=f-\Delta u = f over Ω\Omega. Apply the Divergence Theorem to convert ∫ΩΔu dx\int_\Omega \Delta u\,dx into a boundary integral involving ∂νu\partial_\nu u.
View Solution
Solution to question 1: ∫Ωf dx=−∫ΩΔu dx=−∫∂Ω∂νu dS=−∫∂Ωh dS \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: ∫Ωf dx+∫∂Ωh dS=0 \boxed{\int_\Omega f\,dx + \int_{\partial\Omega} h\,dS = 0} Solution to question 2: If −Δu=f-\Delta u = f, then −Δ(u+C)=−Δu=f-\Delta(u+C) = -\Delta u = f and ∂ν(u+C)=∂νu=h\partial_\nu(u+C) = \partial_\nu u = h, so u+Cu+C is also a solution. To ensure uniqueness, impose a normalization, e.g., ∫Ωu dx=0\displaystyle\int_\Omega u\,dx = 0, which fixes the additive constant.

Problem 10: Mixed Dirichlet–Neumann Problem on a Rectangle

Hard

On the rectangle Ω=(0,L)×(0,H)\Omega=(0,L)\times(0,H), solve the following mixed problem using separation of variables and superposition:

−Δu=0 in Ω -\Delta u = 0 \text{ in } \Omega

with boundary conditions: u(0,y)=0u(0,y) = 0, u(L,y)=0u(L,y) = 0, u(x,0)=0u(x,0) = 0, and ∂yu(x,H)=g(x)\partial_y u(x,H) = g(x).

  1. Seek separated solutions and derive the eigenfunctions satisfying the three homogeneous conditions.
  2. Expand g(x)g(x) in the appropriate eigenfunction basis and write the formal series solution u(x,y)u(x,y).
Hint
Set u(x,y)=X(x)Y(y)u(x,y)=X(x)Y(y). The Dirichlet conditions on x=0x=0, x=Lx=L, and y=0y=0 give Xn(x)=sin⁡(nπx/L)X_n(x)=\sin(n\pi x/L) and Yn(0)=0Y_n(0)=0. The yy-ODE gives Yn(y)=sinh⁡(nπy/L)Y_n(y) = \sinh(n\pi y/L). Apply the Neumann condition at y=Hy=H to find coefficients.
View Solution
Solution to question 1: Separating −Δu=0-\Delta u = 0 with u=X(x)Y(y)u=X(x)Y(y): X′′+λX=0,X(0)=X(L)=0⇒Xn(x)=sin⁡ ⁣(nπxL),  λn=(nπL)2 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′′−λnY=0,Yn(0)=0⇒Yn(y)=sinh⁡ ⁣(nπyL) 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)=∑n=1∞Ansin⁡ ⁣(nπxL)sinh⁡ ⁣(nπyL) 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 ∂yu(x,H)=g(x)\partial_y u(x,H) = g(x) gives: ∑n=1∞AnnπLcosh⁡ ⁣(nπHL)sin⁡ ⁣(nπxL)=g(x) \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⁡(mπx/L)\sin(m\pi x/L) and integrating over (0,L)(0,L): An=2nπcosh⁡(nπH/L)∫0Lg(x)sin⁡ ⁣(nπxL)dx 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 Ω⊂Rn\Omega\subset\mathbb{R}^n be a bounded domain with smooth boundary, and consider: −Δu=f in Ω,u=0 on ∂Ω -\Delta u = f \text{ in } \Omega, \quad u = 0 \text{ on } \partial\Omega with f∈L2(Ω)f\in L^2(\Omega).

  1. Multiply the equation by a test function v∈H01(Ω)v \in H^1_0(\Omega), integrate over Ω\Omega, and apply Green’s first identity to derive the weak formulation: find u∈H01(Ω)u\in H^1_0(\Omega) such that a(u,v)=F(v)a(u,v) = F(v) for all v∈H01(Ω)v\in H^1_0(\Omega). Identify the bilinear form a(⋅,⋅)a(\cdot,\cdot) and the linear functional FF.
  2. State the Lax–Milgram theorem conditions that guarantee a unique weak solution, and verify them for this bilinear form using the Poincaré inequality.
Hint
After multiplying by vv and integrating, apply: ∫Ω(−Δu)v dx=∫Ω∇u⋅∇v dx−∫∂Ω(∂νu)v dS\int_\Omega (-\Delta u)v\,dx = \int_\Omega\nabla u\cdot\nabla v\,dx - \int_{\partial\Omega}(\partial_\nu u)v\,dS. Since v=0v=0 on ∂Ω\partial\Omega, the boundary term vanishes. Lax–Milgram requires continuity and coercivity of aa.
View Solution
Solution to question 1: Multiply −Δu=f-\Delta u = f by v∈H01(Ω)v\in H^1_0(\Omega) and integrate: ∫Ω(−Δu)v dx=∫Ωfv dx \int_\Omega (-\Delta u)v\,dx = \int_\Omega fv\,dx By Green’s first identity (with v=0v=0 on ∂Ω\partial\Omega): ∫Ω∇u⋅∇v dx=∫Ωfv dx \int_\Omega \nabla u\cdot\nabla v\,dx = \int_\Omega fv\,dx The weak formulation is: find u∈H01(Ω)u\in H^1_0(\Omega) such that a(u,v)=F(v)∀ v∈H01(Ω) a(u,v) = F(v)\quad\forall\,v\in H^1_0(\Omega) with a(u,v)=∫Ω∇u⋅∇v dxa(u,v) = \displaystyle\int_\Omega\nabla u\cdot\nabla v\,dx and F(v)=∫Ωfv dxF(v)=\displaystyle\int_\Omega fv\,dx. Solution to question 2: Lax–Milgram requires: (i) continuity: ∣a(u,v)∣≤∥∇u∥L2∥∇v∥L2≤∥u∥H1∥v∥H1|a(u,v)|\leq\|\nabla u\|_{L^2}\|\nabla v\|_{L^2}\leq\|u\|_{H^1}\|v\|_{H^1} ✓; (ii) coercivity: a(u,u)=∥∇u∥L22≥C∥u∥H12a(u,u)=\|\nabla u\|^2_{L^2}\geq C\|u\|^2_{H^1} by the Poincaré–Friedrichs inequality ∥u∥L2≤CP∥∇u∥L2\|u\|_{L^2}\leq C_P\|\nabla u\|_{L^2} for u∈H01(Ω)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: −Δu=f in Ω,u=g on ∂Ω -\Delta u = f \text{ in } \Omega, \quad u = g \text{ on } \partial\Omega where f∈L2(Ω)f\in L^2(\Omega) and g∈H1(Ω)g\in H^1(\Omega) (understood as a trace).

  1. Introduce the lifting g~∈H1(Ω)\tilde{g}\in H^1(\Omega) with g~∣∂Ω=g\tilde{g}|_{\partial\Omega} = g, define w=u−g~w = u - \tilde{g}, and write the equation satisfied by ww with homogeneous Dirichlet data.
  2. Write the weak formulation for ww and invoke the Lax–Milgram theorem to conclude existence and uniqueness of the weak solution uu.
Hint
Substituting u=w+g~u = w + \tilde{g} into the equation gives −Δw=f+Δg~-\Delta w = f + \Delta\tilde{g}, with w=0w=0 on ∂Ω\partial\Omega. The right-hand side f+Δg~f + \Delta\tilde{g} defines a new source in H−1(Ω)H^{-1}(\Omega), to which Lax–Milgram applies.
View Solution
Solution to question 1: Set w=u−g~w = u - \tilde{g}. Then −Δw=−Δu+Δg~=f+Δg~-\Delta w = -\Delta u + \Delta\tilde{g} = f + \Delta\tilde{g} in Ω\Omega, and w=u−g~=g−g=0w = u - \tilde{g} = g - g = 0 on ∂Ω\partial\Omega. The new source is f~=f+Δg~∈H−1(Ω)\tilde{f} = f + \Delta\tilde{g}\in H^{-1}(\Omega). Solution to question 2: The weak formulation for w∈H01(Ω)w\in H^1_0(\Omega) is: ∫Ω∇w⋅∇v dx=∫Ωfv dx−∫Ω∇g~⋅∇v dx∀ v∈H01(Ω) \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 H01(Ω)H^1_0(\Omega) (by Cauchy–Schwarz). Since a(w,v)=∫Ω∇w⋅∇v dxa(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∈H01(Ω)w\in H^1_0(\Omega), and thus a unique weak solution u=w+g~∈H1(Ω)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 Ω⊂Rn\Omega\subset\mathbb{R}^n be a bounded connected domain, and let u∈C2(Ω)∩C(Ω‾)u\in C^2(\Omega)\cap C(\overline{\Omega}) satisfy −Δu≤0-\Delta u \leq 0 in Ω\Omega (i.e., uu is subharmonic).

  1. State the weak maximum principle for subharmonic functions and prove it by contradiction: assume the maximum is achieved at an interior point x0x_0 and derive a contradiction using the second-order condition Δu(x0)≥0\Delta u(x_0) \geq 0.
  2. Use the maximum principle to prove: if u1,u2∈C2(Ω)∩C(Ω‾)u_1, u_2\in C^2(\Omega)\cap C(\overline{\Omega}) both solve −Δu=f-\Delta u = f in Ω\Omega with u1≤u2u_1 \leq u_2 on ∂Ω\partial\Omega, then u1≤u2u_1\leq u_2 throughout Ω‾\overline{\Omega}.
Hint
For part 1: if uu attains its maximum at an interior point x0x_0, then Δu(x0)≤0\Delta u(x_0)\leq 0 by calculus, but −Δu≤0-\Delta u\leq 0 gives Δu≥0\Delta u\geq 0. Combine. For part 2: let w=u2−u1w = u_2 - u_1 and apply the result of part 1 to ww.
View Solution
Solution to question 1: Weak Maximum Principle: max⁡Ω‾u=max⁡∂Ωu\max_{\overline{\Omega}} u = \max_{\partial\Omega} u.
Proof: Suppose for contradiction that uu attains its global maximum at an interior point x0∈Ωx_0\in\Omega. Then by the second-order necessary condition for a maximum: ∂xixiu(x0)≤0\partial_{x_i x_i}u(x_0)\leq 0 for each ii, so Δu(x0)≤0\Delta u(x_0)\leq 0. But −Δu≤0-\Delta u\leq 0 means Δu(x0)≥0\Delta u(x_0)\geq 0. Hence Δu(x0)=0\Delta u(x_0) = 0. A more careful argument (using the mean value inequality for subharmonic functions) shows uu must be constant in a neighborhood of x0x_0, and by connectedness, u≡constu \equiv \text{const} on Ω\Omega, contradicting a strict interior maximum. Thus max⁡Ω‾u=max⁡∂Ωu\max_{\overline{\Omega}}u = \max_{\partial\Omega}u. Solution to question 2: Let w=u2−u1w = u_2 - u_1. Then −Δw=−Δu2+Δu1=f−f=0-\Delta w = -\Delta u_2 + \Delta u_1 = f - f = 0 in Ω\Omega, so ww is harmonic, hence both subharmonic and superharmonic. By the maximum principle applied to ww: max⁡Ω‾w=max⁡∂Ωw≤0 \max_{\overline{\Omega}} w = \max_{\partial\Omega} w \leq 0 since u1≤u2u_1 \leq u_2 on ∂Ω\partial\Omega gives w≤0w \leq 0 on ∂Ω\partial\Omega. Therefore w=u2−u1≤0w = u_2 - u_1 \leq 0 on Ω‾\overline{\Omega}, i.e., u1≤u2u_1\leq u_2 throughout Ω‾\overline{\Omega}.