22.10 More general eigenfunctions

In the simplest problems, the eigenfunctions are sines and cosines. That includes the examples so far. But it is quite easy to get different eigenfunctions. In this example, they will turn out to be products of sines and exponentials.

22.10.1 The physical problem

Find the unsteady temperature distribution in the moving bar below for arbitrary position and time if the initial distribution at time zero and the temperatures of the ends are known.

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22.10.2 The mathematical problem

$$\hbox{\epsffile{svbm1.eps}}$$

• Finite domain $\bar\Omega$: 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $\ell$
• Unknown temperature $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $u(x,t)$
• Parabolic
• One initial condition
• Two Dirichlet boundary conditions
• Constant $\kappa$

Try separation of variables:

$$\sum_n C_n(t) X_n(x)$$

22.10.3 Step 0: Fix the boundary conditions

The $x$-boundary conditions are inhomogeneous:

$$u(0,t)=g_0(t) \qquad u(\ell,t)=g_1(t)$$

So we try finding a $u_0$ satisfying these boundary conditions:

$$u_0(0,t)=g_0(t) \qquad u_0(\ell,t)=g_1(t)$$

A linear expression works:

$$u_0 = A(t) + B(t) x$$

$$A(t) = g_0(t) \qquad A(t) + B(t) \ell = g_1(t)$$

This can be solved to find

$$\fbox{$\displaystyle u_0(x,t) = g_0(t) + \frac{g_1(t)-g_0(t)}{\ell} x $}$$

To get rid of the inhomogeneous boundary conditions, we subtract $u_0$ from $u$. That will produce homogeneous boundary conditions for the remainder $v$ $\vphantom0\raisebox{1.5pt}{$=$}$ $u-u_0$. Indeed, if you plug $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $u_0+v$ into the boundary conditions, you get

$$u_0(0,t)+v(0,t)=g_0(t) \qquad u_0(\ell,t)+v(\ell,t)=g_1(t)$$

And since $u_0$ satisfies the inhomogeneous boundary conditions, that becomes

$$v(0,t)=0 \qquad v(\ell,t)=0$$

Substitute $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $u_0+v$ into the partial differential equation $u_t$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\kappa u_{xx} + b u_x + cu$ to get

$$v_t = \kappa v_{xx} + b v_x + c v + q$$

where

$$\fbox{$\displaystyle q(x,t) = -g_0'(t) - \frac{g_1'(t)-g... ..._0(t)}{\ell} + c g_0(t) + c \frac{g_1(t)-g_0(t)}{\ell} x $}$$

Substitute $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $u_0+v$ into the initial condition $u(x,0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f(x)$:

$$v(x,0) = \bar f(x)$$

$$\fbox{$\displaystyle \bar f(x) = f(x) - g_0(0) - \frac{g_1(0)-g_0(0)}{\ell} x$}$$

The problem for $v$ is therefor:

$$\hbox{\epsffile{svbm2.eps}}$$

22.10.4 Step 1: Find the eigenfunctions

Substitute $v$ $\vphantom0\raisebox{1.5pt}{$=$}$ $T(t) X(x)$ into the homogeneous partial differential equation $v_t$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\kappa v_{xx} + bv_x+cv$:

$$T'X = \kappa T X'' + b T X' + c T X$$

Separate:

$$\frac{T'}{T} = \kappa \frac{X''}{X} + b \frac{X'}{X} + c = \hbox{ constant } = - \lambda$$

The Sturm-Liouville problem for $X$ is now:

$$- \kappa X'' - b X' - c X = \lambda X \qquad X(0) = 0 \qquad X(\ell)=0$$

This is a constant coefficient ordinary differential equation, with a characteristic polynomial:

$$\kappa k^2 + b k + (c + \lambda) = 0$$

The fundamentally different cases are now two real roots (discriminant positive), a double root (discriminant zero), and two complex conjugate roots (discriminant negative.) We do each in turn.

2. Case $b^2 - 4\kappa(c+\lambda)$ $\raisebox{.3pt}{$>$}$ 0:

   Roots $k_1$ and $k_2$ real and distinct:
   

   $$X = A e^{k_1 x} + B e^{k_2 x}$$
   
   

   Boundary conditions:
   

   $$X(0) = 0 = A + B \quad\quad\Rightarrow\quad\quad B = -A$$
   
   
   
   
   $$X(\ell) = 0 = A \left(e^{k_1\ell}-e^{k_2\ell}\right) = 0$$
   
   
   No nontrivial solutions since the roots are different.

3. Case $b^2 - 4\kappa(c+\lambda)$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0:

   Since $k_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $k_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $k$:
   

   $$X = A e^{k x} + B x e^{k x}$$
   
   

   Boundary conditions:
   

   $$X(0) = 0 = A \qquad X(\ell) = 0 = B \ell e^{k\ell}$$
   
   
   No nontrivial solutions.

4. Case $b^2 - 4\kappa(c+\lambda)$ $\raisebox{.3pt}{$<$}$ 0:

   For convenience, we will write the roots of the characteristic polynomial more concisely as:
   

   $$k_1 = -\mu + {\rm i} \omega \quad k_2 = -\mu - {\rm i} \omega$$
   
   
   where according to the solution of the quadratic
   
   $$\mu = \frac{b}{2\kappa} \qquad \omega=\frac{\sqrt{4\kappa(c+\lambda)-b^2}}{2 \kappa}$$
   
   
   Since it can be confusing to have too many variables representing the same thing, let's agree that $\mu$ is our ``representative'' for $b$, and $\omega$ our ``representative'' for $\lambda$. In terms of these representatives, the solution is, after clean-up,
   
   $$X = e^{-\mu x}\left(A \cos(\omega x) + B \sin(\omega x)\right)$$
   
   

   Boundary conditions:
   

   $$X(0) = 0 = A \qquad X(\ell) = 0 = e^{-\mu \ell} B \sin(\omega \ell)$$
   
   

   Nontrivial solutions $B\ne 0$ can only occur if
   

   $$\sin(\omega \ell) = 0 \quad\quad\Rightarrow\quad\quad \omega_n = n \pi/\ell \quad (n = 1,2,\ldots)$$
   
   
   which gives us our eigenvalues, by substituting in for $\omega$:
   
   $$\lambda_n =\frac{\kappa n^2 \pi^2}{\ell^2} + \frac{b^2}{4\kappa} - c \quad (n=1,2,3,...)$$
   
   

   Also, choosing each $B$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1:
   

   $$X_n = e^{-\mu x} \sin\left(n \pi x/\ell\right) \quad (n=1,2,3,...)$$
   
   

22.10.5 Step 2: Solve the problem

Expand all variables in the problem for $v$ in a Fourier series:

$$\hbox{\epsffile{svbm2.eps}}$$

$$v = \sum_{n=1}^\infty v_n(t) X_n(x) \quad \bar f = \sum_... ...y \bar f_n X_n(x) \quad q = \sum_{n=1}^\infty q_n(t) X_n(x)$$

We want to first find the Fourier coefficients of the known functions $\bar f$ and $q$. Unfortunately, the ordinary differential equation found in the previous section,

$$- \kappa X'' - b X' - c X = \lambda X$$

is not in standard Sturm-Liouville form: the derivative of the first, $X''$, coefficient, $-\kappa$, is zero, not $-b$. Let's try to make it OK by multiplying the entire equation by a factor, which will then be our $\bar r$.

$$- \bar {\rm r} \kappa X'' - \bar {\rm r} b X' - \bar {\rm r} c X = \lambda \bar {\rm r} X$$

We want that the second coefficient is the derivative of the first:

$$\bar {\rm r} b = \frac{{\rm d}}{{\rm d}x} \left(\bar {\rm r} \kappa\right)$$

This is a simple ordinary differential equation for the $\bar {\rm r}$ we are trying to find, and a valid solution is:

$$\bar {\rm r} = e^{bx/\kappa} = e^{2\mu x}$$

Having found $\bar {\rm r}$, we can write the orthogonality relationships for the generalized Fourier coefficients of $\bar f$ and $q$ (remember that $X_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{-\mu x}\sin(n\pi x/\ell)$):

$$\bar f_n = \frac {\int_{x=0}^{\ell} e^{\mu x} \bar f(x)\... ...,\rm d}x} {\int_{x=0}^{\ell} \sin^2(n\pi x/\ell){ \rm d}x}$$

$$q_n(t) = \frac {\int_{x=0}^{\ell}e^{\mu x} q(x,t)\sin(n\... ...,\rm d}x} {\int_{x=0}^{\ell} \sin^2(n\pi x/\ell){ \rm d}x}$$

The integrals in the bottoms equal $\ell/2$.

Expand the partial differential equation $v_t$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\kappa v_{xx} + b v_x + c v + q$ in a generalized Fourier series:

$$\begin{array}{l} \displaystyle \sum_{n=1}^\infty \dot ... ...isplaystyle + \sum_{n=1}^\infty q_n(t) X_n(x) \end{array}$$

Because of the choice of the $X_n$, $\kappa X''+bX'+cX$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\lambda X$:

$$\sum_{n=1}^\infty \dot v_n(t) X_n(x) = - \sum_{n=1}^\infty \lambda_n v_n(t) X_n(x) + \sum_{n=1}^\infty q_n(t) X_n(x)$$

So, the ordinary differential equation for the generalized Fourier coefficients of $v$ becomes:

$$\dot v_n(t) + \lambda_n v_n(t) = q_n(t)$$

Expand the initial condition $v(x,0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\bar f(x)$ in a generalized Fourier series:

$$\sum_{n=1}^\infty v_n(0) X_n(x) = \sum_{n=1}^\infty \bar f_n X_n(x)$$

so

$$v_n(0) = \bar f_n$$

Solve this ordinary differential equation and initial condition for $v_n$:

Homogeneous equation:

$$v_{nh} = A_n e^{-\lambda_n t}$$

Inhomogeneous equation:

$$A_n' e^{-\lambda_n t} + 0 = q_n(t)$$

$$A_n = \int_{\tau=0}^t q_n(\tau) e^{\lambda_n \tau} { \rm d}\tau + A_{n0}$$

$$v_n = A_n e^{-\lambda_n t}$$

$$v_n = \int_{\tau=0}^t q_n(\tau) e^{- \lambda_n(t - \tau)} { \rm d}\tau + A_{n0} e^{-\lambda_n t}$$

Initial condition: $A_{n0}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\bar f_n$.

$$v_n = \int_{\tau=0}^t q_n(\tau) e^{- \lambda_n(t - \tau)} { \rm d}\tau + \bar f_n e^{-\lambda_n t}$$

22.10.6 Summary of the solution

Total solution:

$$\mu = \frac{b}{2\kappa} \qquad \lambda_n =\frac{\kappa n... ...l^2} + \lambda_0 \qquad \lambda_0 = \frac{b^2}{4\kappa} - c$$

$$\bar f(x) = f(x) - g_0(0) - \frac{g_1(0)-g_0(0)}{\ell} x$$

$$\bar f_n = \frac{2}{\ell} \int_{x=0}^{\ell} \bar f(x) e^{\mu x}\sin(n\pi x/\ell){ \rm d}x$$

$$q(x,t) = -g_0'(t) - \frac{g_1'(t)-g_0'(t)}{\ell} x + b \... ...ell} + c \left(g_0(t) + \frac{g_1(t)-g_0(t)}{\ell} x\right)$$

$$q_n(t) = \frac{2}{\ell} \int_{x=0}^{\ell} q(x,t) e^{\mu x}\sin(n\pi x/\ell){ \rm d}x$$

$$\begin{array}{l} \displaystyle u_{\strut} = g_0(t) + \... ..._n t} \right] e^{- \mu x} \sin(n\pi x/\ell) \end{array}$$

22.10.7 An alternative procedure

Define a new unknown $w$ by $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $w e^{-\alpha x -\beta t}$. Put this in the partial differential equation for $u$ and choose $\alpha$ and $\beta$ so that the $w_x$ and $w$ terms drop out. This requires:

$$u = w e^{-\mu x - \lambda_0 t}$$

Then:

$$w_t = \kappa w_{xx} \quad w(x,0)= e^{\mu x} f(x) \quad w... ...} g_0(t) \quad w(\ell,t) = e^{\mu\ell + \lambda_0 t} g_1(t)$$

No fun! Note that the generalized Fourier series coefficients for $u$ become normal Fourier coefficients for $w$.