Terminology — Question 3

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Question 3

Let u(x,t)u(x,t) solve ut=κuxxu_t=\kappa u_{xx} on 0<x<L0<x<L, 0<t<T0<t<T, where κ,L,T>0\kappa,L,T>0. For this problem, initial data are prescribed at t=0t=0; spatial boundary data are prescribed at x=0,Lx=0,L. The outward normal derivative is ∂nu=nxux\partial_n u=n_xu_x, with nx=−1n_x=-1 on the left and nx=1n_x=1 on the right. Consider u(x,0)=f(x),u(0,t)=A(t),κ∂nu(L,t)+hu(L,t)=r(t),h>0.u(x,0)=f(x),\qquad u(0,t)=A(t),\qquad \kappa\partial_n u(L,t)+h u(L,t)=r(t),\quad h>0. Dirichlet data prescribe uu, Neumann data prescribe ∂nu\partial_nu, and Robin data prescribe a linear combination with nonzero coefficients of both.

Tasks

  1. Classify each condition, and identify the portions of the space–time rectangle on which they are prescribed. Is t=Tt=T a spatial boundary?

  2. Rewrite the right condition with uxu_x. Write the corresponding coordinate form of κ∂nu+hu=r\kappa\partial_nu+h u=r if it were imposed at the left endpoint instead.

  3. Classify the spatial boundary operator when h=0h=0, and decide exactly when the displayed spatial data are homogeneous. Distinguish this from homogeneity of the PDE and of the initial data.

  4. Verify the manufactured field u*(x,t)=x2+2κtu_*(x,t)=x^2+2\kappa t. Find f,A,rf,A,r that make it solve the full displayed problem, and illustrate where each datum acts.

Original worksheet page 1: question and worked solution for 9-3-003
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Question 3 – Solution

Strategy. Identify both the operator being prescribed and the geometric location of the data.

Step 1: Locate and name the data. The bottom edge t=0t=0 carries initial displacement/temperature data f(x)f(x). The left edge carries Dirichlet data A(t)A(t); the right carries Robin data r(t)r(t). The top t=Tt=T is a final-time slice, not a spatial boundary. No data are prescribed there in this initial-boundary-value problem.

Step 2: Keep the outward signs. At the right, the condition is κux(L,t)+hu(L,t)=r(t)\kappa u_x(L,t)+h u(L,t)=r(t). At the left, the analogous outward-normal condition would be −κux(0,t)+hu(0,t)=r(t)-\kappa u_x(0,t)+h u(0,t)=r(t). Changing the endpoint changes the sign of the normal derivative, not the sign of the prescribed value by convention.

Step 3: Classify homogeneity at each level. When h=0h=0, the right condition is Neumann, since κ>0\kappa>0. The spatial data are homogeneous exactly when A≡0A\equiv 0 and r≡0r\equiv 0. The PDE is already linear homogeneous; initial data are homogeneous exactly when f≡0f\equiv 0. These are three separate assertions. A nonzero Robin coefficient hh does not itself make the condition nonhomogeneous.

Step 4: Verify all parts of a full problem. Here u*,t=2κu_{*,t}=2\kappa and u*,xx=2u_{*,xx}=2, so the PDE holds. Evaluation gives f(x)=x2,A(t)=2κt,r(t)=2κL+h(L2+2κt).\boxed{f(x)=x^2,\quad A(t)=2\kappa t,\quad r(t)=2\kappa L+h(L^2+2\kappa t).} These data also agree at their shared corners. The diagram locates the data; it is a space–time domain, not a graph of the solution value.

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 9-3-003

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