21. Solve the Neumann problem for a rectangle: (∂^2 u)/(∂...

21. Solve the Neumann problem for a rectangle: $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$, $0 < x < a$, $0 < y < b$ $\frac{\partial u}{\partial y}|_{y=0} = 0$, $\frac{\partial u}{\partial y}|_{y=b} = 0$, $0 < x < a$ $\frac{\partial u}{\partial x}|_{x=0} = 0$, $\frac{\partial u}{\partial x}|_{x=a} = g(y)$, $0 < y < b$. (a) Explain why a necessary condition for a solution $u$ to exist is that $g$ satisfy $\int_0^b g(y)dy = 0$. This is sometimes called a compatibility condition. Do some extra reading and explain the compatibility condition on physical grounds. (b) If $u$ is a solution of the BVP, explain why $u + c$, where $c$ is an arbitrary constant, is also a solution.

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00:01 In part a to apply the method of the separation of variables to the given pd we can assume that the solutions can be written as a product of the two functions one depending only on x and other only on t so u of x comma t is equal to x of x into t of t substituting this into the pd we get 1 by t into dt by dt which is equal to 3 the root of 1 plus x square by x into dx by dx since the left hand side depends only on t or the right hand side depends only on x they must be equal to a constant say minus lambda square so 1 by t dt by dt is equal to minus lambda square then 3 into root of 1 plus x square by x dx with respect to its small dx is equal to minus lambda square these are the ordinary differential equation implied by the method of separation of variables for part b solving the first od for t of t we get t of t is equal to c1 e power minus lambda square t solving the second ode for x of x has more involved.

01:13 We can rewrite it as x of t sorry dx by dx is equal to minus lambda square into 3 of 1 plus x square into x this is a purple first order od which we can solve by separation of variables so dx by x is equal to minus lambda square by 3 into root of 1 plus x square into dx the integrating on both sides we get i n of x is equal to minus lambda cube by 3 i n of 1 plus x square plus c2 and x of x is equal to c3 into 1 plus x square the whole power minus lambda square by 3 the general form of the solution is therefore u of x comma t is equal to c3 e power minus lambda square t into 1 plus x square into minus lambda square by 3 now od for x of x is a term liable problem so this is part c it is a self adjoint operator with eigenvalues that form an orthonormal basis of the function subspace the eigenvalues are the constants lambda square that satisfy the boundary conditions in our case the boundary conditions are u of minus 1 comma t which is equal to u of 1 comma t which is equal to 0 substituting the general form of the solution we get c3 e power minus lambda square t into 1 plus minus 1 the whole square the whole power minus lambda square by 3 which is equal to 0 and c3 e power minus lambda square t of 1 plus 1 minus lambda square by 3 which is equal to 0 since e power minus lambda square t is never 0 you must have 1 plus 1 minus lambda square by 3 is equal to 0 this implies that lambda square is not real since a real number raised to a real power is always positive therefore the eigenvalues are complex and corresponding eigenfunctions are complex valued thank you and part d since the eigenvalues are complex then the solution was complex exponentials with gk or grow exponentially as t increases therefore the behavior of the solution for t tends to plus infinity depends on the sign of the real part of the eigenvalues if the real part is negative the solution decays to 0 as t increases if the real part is positive solution grows without bound as t increases if the real part is 0 solution oscillates with constant amplitude as t increases the exact behavior of the solution depends on the specific value of the eigenvalues which can be determined by solving the boundary value problem for x of x...

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