MATH529: L09 Nonhomogeneous BVPs

Overview

• Definition

• Substitutions and superposition

Nonhomogeneous PDE BVPs

• A boundary-value problem for a partial differential equations is nonhomogeneous if either the PDE or the boundary conditions are nonhomogeneous:

  $k{u}_{xx}+F\left(x\right)=u_t$ for $0<x<L$, $t>0$, $u(0,t)=u_0$, $u(L,t)=u_1$, $t>0$, $u(x,0)=f\left(x\right)$, $0<x<L$ with $k>0$

• Separation of variables does not work when applied directly to the full problem. For linear PDEs a solution is obtained by substitution $u(x,t)=v(x,t)+\psi \left(x\right)$, and superposition of solution from two problems

  1. $k{\psi }^{\text{'}\text{'}}+F\left(x\right)=0$, $0<x<L$, $\psi \left(0\right)=u_0$, $\psi \left(L\right)=u_1$ (steady-state eq)

  2. $k{v}_{xx}=v_t$, $0<x<L$, $t>0$, $v(0,t)=0$, $v(L,t)=0$, $v(x,0)=f\left(x\right)-\psi \left(x\right)$

  (transient eq)

Nonhomogeneous PDE BVP example

• Constant heat source: $k{u}_{xx}+r=u_t$ for $0<x<L$, $t>0$, $u(0,t)=u_0$, $u(L,t)=u_1$, $t>0$, $u(x,0)=f\left(x\right)$, $0<x<L$ with $k>0$

  1. $k{\psi }^{\text{'}\text{'}}+r=0$, $0<x<L$, $\psi \left(0\right)=u_0$, $\psi \left(L\right)=u_1$ (steady-state eq)

     ${\displaystyle {\psi }^{\text{'}\text{'}}=-\frac{r}{k}\Rightarrow \psi \left(x\right)=-\frac{r}{k}{x}^2+Ax+u_0,\psi \left(L\right)=u_1=-\frac{r}{k}{L}^2+AL+u_0\Rightarrow }$ ${\displaystyle \frac{u_1-u_0}{L}=-\frac{r}{k}L+A\Rightarrow A=\frac{u_1-u_0}{L}+\frac{r}{k}L\Rightarrow }$ ${\displaystyle \psi \left(x\right)=-\frac{r}{k}{x}^2+(\frac{u_1-u_0}{L}+\frac{r}{k}L)x+u_0}$

  2. $k{v}_{xx}=v_t$, $0<x<L$, $t>0$, $v(0,t)=0$, $v(L,t)=0$, $v(x,0)=f\left(x\right)-\psi \left(x\right)$

(cont)

• Steady-state solution

  ${\displaystyle \psi \left(x\right)=-\frac{r}{k}{x}^2+(\frac{u_1-u_0}{L}+\frac{r}{k}L)x+u_0}$

• Transient problem

  $k{v}_{xx}=v_t$, $0<x<L$, $t>0$, $v(0,t)=0$, $v(L,t)=0$, $v(x,0)=f\left(x\right)-\psi \left(x\right)$

  ${\displaystyle v(x,t)=X\left(x\right)T\left(t\right)\Rightarrow k\frac{X^{\text{'}\text{'}}}{X}=\frac{T^{\text{'}}}{T}\Rightarrow }$ ${\displaystyle v(x,t)=\sum _{n=1}^{\infty }{b}_n\sin ({\alpha }_{n}x)\exp [-k{\alpha }_n^{2}t],{\alpha }_n=\frac{n\pi }{L}}$ ${\displaystyle b_n=(f-\psi ,\sin \left({\alpha }_{n}x\right))=\frac{2}{L}\underset{0}{\overset{L}{\int }}[f\left(x\right)-\psi \left(x\right)]\sin \left({\alpha }_{n}x\right)dx}$

Nonhomogeneous PDE BVP example. Time-dependent BCs

• $u_t=u_{xx}$, $0<x<1$, $t>0$, $u(0,t)=\cos t$, $u(1,t)=0$, $u(x,0)=0$

  ${\displaystyle u(x,t)=X\left(x\right)T\left(t\right)\Rightarrow \frac{T^{\text{'}}}{T}=\frac{X^{\text{'}\text{'}}}{X}=-a^2\Rightarrow }$ ${\displaystyle X\left(x\right)=A\cos (ax)+B\sin (ax)\Rightarrow }$ ${\displaystyle X\left(0\right)=A=\cos t✠(\mathrm{separation}\mathrm{of}\mathrm{variables}\mathrm{fails})}$

• Use superposition and substitution

  ${\displaystyle u(x,t)=v(x,t)+\psi (x,t)}$

Define two problems A,B

• Problem: $u_t=u_{xx}$, $0<x<1$, $t>0$, $u(0,t)=\cos t$, $u(1,t)=0$, $u(x,0)=0$

• Superposition: $u(x,t)=v(x,t)+\psi (x,t)$

  ${\displaystyle v_t+{\psi }_t=v_{xx}+{\psi }_{xx}}$ ${\displaystyle v(x,0)=-\psi (x,0),\psi (0,t)=u_0\left(t\right),\psi (L,t)=u_1\left(t\right)}$

• Define problems for $\psi ,v$:

  1. $v_t=v_{xx}$

  2. ${\psi }_t={\psi }_{xx}$

• Alternative splittings: $v_t={\psi }_{xx}$, ${\psi }_t=v_{xx}$; $v_t=0$, ${\psi }_t=v_{xx}+{\psi }_{xx}$; $v_t=-{\psi }_t$, $v_{xx}=-{\psi }_{xx}$; an infinity of possible splittings. Seek a solvable, simple choice

(cont)

• Problem: $u_t=u_{xx}$, $0<x<1$, $t>0$, $u(0,t)=\cos t$, $u(1,t)=0$, $u(x,0)=0$

• Superposition: $u(x,t)=v(x,t)+\psi (x,t)$

  ${\displaystyle v_t+{\psi }_t=v_{xx}+{\psi }_{xx}}$ ${\displaystyle v(x,0)=-\psi (x,0),\psi (0,t)=u_0\left(t\right),\psi (1,t)=u_1\left(t\right)}$

• Choose a simple form for $\psi (x,t)=u_0\left(t\right)+x[u_1\left(t\right)-u_0\left(t\right)]$, $\psi (0,t)=u_0\left(t\right)=\cos t$, $\psi (1,t)=u_1\left(t\right)=0$, $\psi (x,t)=(1-x)\cos t$

  ${\displaystyle v_t+(x-1)\sin t=v_{xx}}$ ${\displaystyle v(x,0)=x-1,v(x,1)=0}$ ${\displaystyle v(x,t)=X\left(x\right)T\left(t\right)\Rightarrow \frac{T^{\text{'}}}{T}+\frac{(x-1)\sin t}{XT}=\frac{X^{\text{'}\text{'}}}{X},✠.}$

(cont)

• Another approach. Recall inhomogeneous ODE method. To solve

  ${\displaystyle m{y}^{\text{'}\text{'}}+ky=\cos (\Omega t)}$

  first solve homogeneous equation to obtain $y\left(t\right)=A\cos (\omega t)+B\sin (\omega t)$ and then a particular solution of the inhomogeneous equation is sought by superposition of eigenfunctions of $ℒy=-(k/m)y$ with the forcing term ($ℒ=d^2/d{t}^2$, e.g., $\cos (\omega t)$, $\sin \left(\omega t\right)$, ${\omega }^2=k/m$). Obtain solution

  ${\displaystyle y\left(t\right)=A\cos (\omega t)+B\sin (\omega t)+\frac{\cos (\Omega t)}{m({\omega }^2-{\Omega }^2)}}$

• This suggests looking for a solution in terms of the eigenfunctions of the homogeneous problem

  ${\displaystyle v(x,t)=\sum _{n=1}^{\infty }{v}_n\left(t\right)\sin (n\pi x),(x-1)\sin t=\sum _{n=1}^{\infty }{g}_n\left(t\right)\sin (n\pi x)}$

(cont)

• Obtain

  ${\displaystyle g_n\left(t\right)=2\sin t\underset{0}{\overset{1}{\int }}(x-1)\sin (n\pi x)dx.}$

• Replace in equation

  ${\displaystyle \sum _{n=1}^{\infty }[{\dot{v}}_n-{(n\pi )}^2{v}_n]\sin (n\pi x)=-\sum _{n=1}^{\infty }{g}_n\left(t\right)\sin (n\pi x).}$

• Recall that $\{\sin \left(\pi x\right),\sin \left(2\pi x\right),\dots \}$ are linearly independent, hence coefficients can be identified to obtain the ODEs

  ${\displaystyle {\dot{v}}_n-{(n\pi )}^2{v}_n=-g_n\left(t\right)}$