Definitions, Theorems, Examples, Topics for Asmar 2nd edition Chapter 3: 3.9 only 3.9 Poisson's Equation: The method of Eigenfunction Expansions ==== The Poisson Problem: u_xx + u_yy = f(x,y) u(x,y) given on the boundary of the rectangle. u(x,0)=f1(x), u(x,b)=f2(x) for 0 = double integral of g(x,y)h(x,y) over the rectangle The usual orthogonality trick will find the constants. Then -lambda[m][n] E[m][n] = (4/(ab)) THEOREM. The solution of the Poisson problem is u = u1 + u2, where u1 is the solution of Problem (a) and u2 is the solution of Problem (b). EXAMPLE 1. Solve u_xx + u_yy =1 in a 1x1 square with u=0 on all edges. This is problem (a) above, solved with the modes phi[m][n]. The solution will be completed, by evaluating the constants E[m][n], in the series expansion u(x,y) = double sum of E[m][n] times sin(m Pi x)sin(n Pi y) Rather than use any previous work, we will find the answer for the integral in the formula for E[m][n]: -(m^2+n^2)Pi^2 E[m][n] = 4 double integral of F(x,y) F(x,y)=f(x,y)sin(m Pi x)sin(n Pi y) Rectangle is 0<=x<=1, 0<=y<=1. And f(x,y)=1. This is an iterated integral, equivalent to 2 normal calculus integrations. integral[1] = int(sin(m Pi x),x=0..1) = (-cos(m Pi) + 1)/(m Pi) = (1-(-1)^m)/(m Pi) integral[2] = int(sin(n Pi y),y=0..1) = (-cos(n Pi) + 1)/(n Pi) = (1-(-1)^n)/(n Pi) Then the answer is -(m^2+n^2)Pi^2 E[m][n] = 4 double integral of F(x,y) = 4 integral[1] integral[2] = 4(2)(2)/(mnPi^2) both m,n odd =0 otherwise See the equation display for u(x,y) at the bottom of page 172. Graphs in 3D are made by truncating the double series to a few terms. EXAMPLE 2. Eigenfunction Expansion Method Solve u_xx + u_yy = u +3 on a 1x1 square with u=0 on the 4 edges. PLAN We use u=superposition of Helmholtz eigenfunctions, following EXAMPLE 1. The u defined by this double series with coefficients E[m][n] will satisfy a Poisson equation with rather strange f(x,y). The modes phi[m][n] satisfy a Helmholtz equation u_xx+u_yy=-cu, so the strange function f(x,y) is a superposition of the modes: f(x,y) = u_xx + u_yy = double series of constants times modes The constants are -lambda[m][n] E[m][n] for corresponding modes phi[m][n] Now we start a competition with the Example 2 differential equation u_xx + u_yy = u + 3 Clearly the two equations imply f(x,y) = u + 3 which has a double series on both sides. The plan is collect the double series on the left, leaving 3 on the right, then solve for the constants using orthogonality of the eigenfunctions phi[m][n]. ASMAR does the double series work on page 174. It has 14 summation signs, finally arriving at the double series identity double sum of constants times modes = 3 modes = phi[m][n] constants = (-1-lambda[m][n])E[m][n] By orthogonality, we can compute the constants: a constant = <3,mode>/ This pair of double integrals can be evaluated from the computations in Example 1. Nothing new happened. Then divide the equation above by (-1-lambda[m]n[]) to isolate E[m][n]. The final answer of the mystery coefficients E[m][n] in the superposition of modes for the solution u(x,y) is given by E[m][n] = 0 unless both m,n are odd, in which case E[m][n] = (-48/(mnPi^2))/(1+lambda[m][n]) A ONE DIMENSIONAL EIGENVALUE PROBLEM This is the Sturm-Liouville Problem obtained from the rectangle problem by looking only along the edge 0<=x<=a. u''(x) + lambda u(x) = 0, u(0)=0, u(a)=0. This was solved earlier in Asmar's book, with answer There is a nonzero solution only in case lambda is positive and the square of an integer m. Then the solution is u[m](x) = sin(m Pi x/a). We call lambda[m]=(m Pi/a)^2 the eigenvalue for u[m](x). A SINGLE SERIES SOLUTION OF THE POISSON PROBLEM The idea comes from separation of variables. A double series solution actually looks like u(x,y) = single series (series in y) sin(m Pi x/a) = single series of terms E[m](y) sin(m Pi x/a) where E[m] stands for the series in y This technique is much like stuffing a product solution X(x)Y(y) into the differential equation. Only it is a series, instead. The equation is Poisson's equation u_xx + u_yy = f(x,y) Stuffing in the single series for u, the left side will be a single series, after collecting terms: single series (y-expression)sin(m Pi x/a) = f(x,y) where y-expression = v'' -(m Pi/a)^2v with v=E[m](y). Another competition to be solved. Suppose y is held fixed. Then the y-expression is a constant. Fourier series theory implies it equals the Fourier coefficient of sin(m Pi x/a) for the function x --> f(x,y) [y is not a variable]. The competing equation for the constant is then y-expression=constant=Fourier coefficient or v'' -(m Pi/a)^2 v = (2/a)(integral of f(x,y)sin(m Pi x/a),x=0..a) with boundary conditions v(0)=0 and v(b)=0. This complicated problem is routinely solved in engineering ODE courses using the variation of parameters formula and Euler's method for solving constant coefficient equations (its a 2250 problem). ASMAR gives the solution as equation (21) on page 176. What he didn't say is that Maple could solve this problem, symbolically, just in case the prospect of doing it yourself makes you a bit weak in the knees. Why did it work? Because the homogeneous problem v'' - (m Pi/a)^2 v = 0, v(0)=0, v(b)=0, has only the solution v=0. This is why the non-homogeneous problem above has a unique solution. When it does, you can find it, using the variation of parameters formula's particular solution v_p: v_p(y) = v[1] int(-v[2] F(y)) + v[2] int(v[1] F(y)) where v[1], v[2] is a basis for the solution space of the homogeneous problem v'' - (m Pi/a)^2 v = 0.