Solution. The separation of variables is the same as for Dirichlet boundary conditions done in Chapter 1. We therefore obtain \(u(t,x)=T(t)X(x)\) where

\[ T'=\lambda T,\qquad X''=\lambda X. \]

For the boundary conditions we similarly as in Chapter 1 obtain

\[ X'(0)=0,\qquad X'(L)=0. \]

We therefore consider the ODE boundary value problem

\[ X''=\lambda X,\qquad X'(0)=0,\qquad X'(L)=0. \]

If \(\lambda >0\), then the general solution is

\[ X(x)=A\cosh (\sqrt {\lambda }x)+B\sinh (\sqrt {\lambda }x), \]

which gives

\[ X'(x)=A\sqrt {\lambda }\sinh (\sqrt {\lambda }x)+B\sqrt {\lambda }\cosh (\sqrt {\lambda }x), \]

so that \(X'(0)=B\sqrt {\lambda }\), and from \(X'(0)=0\) we therefore obtain \(B=0\). Then \(X'(x)=A\sqrt {\lambda }\sinh (\sqrt {\lambda }x)\) and therefore \(X'(L)=0\) gives \(A\sqrt {\lambda }\sinh (\sqrt {\lambda }L)\). Since the hyperbolic sine only has zero as a zero, this gives \(A=0\). Hence we only obtain the trivial (zero) solution. Therefore we must have \(\lambda \leq 0\). For \(\lambda =0\) we obtain precisely the constant functions as solutions. For \(\lambda <0\) we write \(\lambda =-\omega ^2\) with \(\omega >0\). Our ODE boundary value problem then is

\[ X''=-\omega ^2 X,\qquad X'(0)=0,\qquad X'(L)=0. \]

The general solution of this ODE is

\[ X(x)=A\cos (\omega x)+B\sin (\omega x), \]

which gives

\[ X'(x)=-A\omega \sin (\omega x)+B\omega \cos (\omega x). \]

Imposing the boundary condition \(X'(0)=0\) gives \(B=0\) and subsequently imposing the boundary condition \(X'(L)=0\) gives \(\sin (\omega L)=0\). We therefore obtain

\[ \omega _k=\frac {k\pi }{L},\qquad k\in \mN , \]

which leads to the solutions (the normalization constant chosen so that \(\int _0^L X_k(x)^2\,dx=1\))

\[ X_k(x)=\sqrt {\frac {2}{L}}\cos \left (\frac {k\pi x}{L}\right ). \]

For \(\lambda =0\) we further obtain (the normalization constant chosen so that \(\int _0^L X_0(x)^2\,dx=1\))

\[ X_0(x)=\sqrt {\frac {1}{L}}. \]

Note that we have

\[ \lambda _k=-\left (\frac {k\pi }{L}\right )^2,\qquad k\in \mN _0. \]

The differential equation for \(T\) gives

\[ T_k(t)=C_k\e ^{\lambda _k t}, \]

so that we obtain the solutions

\[ u_k(t,x)=C_k\e ^{\lambda _k t}X_k(x),\qquad k\in \mN _0. \]

For the solution of the initial value problem we then obtain

\[ u(t,x)=\sum _{k=0}^\infty C_k\e ^{\lambda _k t}X_k(x), \]

where (using orthonormality of the \(X_k\) as in Chapter 1)

\[ C_k=\int _0^L u^0(x)X_k(x)\,dx. \]

□

Solution. Separation of variables as in Chapter 1 now leads to

\[ T''=\lambda T,\qquad X''=\lambda X. \]

The ODE boundary value problem for \(X\) is the same as in Chapter 1 and therefore we obtain

\[ X_k(x)=\sqrt {\frac {2}{L}}\sin \left (\frac {k\pi x}{L}\right ),\qquad \lambda _k=-\left (\frac {k\pi }{L}\right )^2,\qquad k\in \mN . \]

The differential equation for \(T\) then is

\[ T_k''=\lambda _kT_k, \]

which has general solution

\[ T_k=C_k\sin \left (\frac {k\pi t}{L}\right )+D_k\cos \left (\frac {k\pi t}{L}\right ). \]

We therefore obtain the solutions

\[ u_k(t,x)=\left [C_k\sin \left (\frac {k\pi t}{L}\right )+D_k\cos \left (\frac {k\pi t}{L}\right )\right ]X_k(x). \]

From this we form the solution

\[ u(t,x)=\sum _{k=1}^\infty \left [C_k\sin \left (\frac {k\pi t}{L}\right )+D_k\cos \left (\frac {k\pi t}{L}\right )\right ]X_k(x). \]

The initial conditions \(u(0,x)=u^0(x)\) and \(\partial _t u(0,x)=u^1(x)\) then become

\[ \sum _{k=1}^\infty D_kX_k(x)=u^0(x),\qquad \sum _{k=1}^\infty C_k~\frac {k\pi }{L}~X_k(x)=u^1(x). \]

Using orthonormality of the \(X_k\), this gives

\[ D_k=\int _0^L u^0(x)X_k(x)\,dx,\qquad C_k=\frac {L}{k\pi }\int _0^L u^1(x)X_k(x)\,dx. \]

□

Solution. The separation of variables is similar to the heat equation on a rectangle of Chapter 3. The only difference is that we now obtain

\[ T''=\lambda T. \]

As in Chapter 3 we obtain

\[ \lambda _{km}=-\left (\frac {k\pi }{L_1}\right )^2-\left (\frac {m\pi }{L_2}\right )^2, \]

and

\[ X_k(x)=\sqrt {\frac {2}{L_1}}\sin \left (\frac {k\pi x}{L_1}\right ),\qquad \lambda ^X_k=-\left (\frac {k\pi }{L_1}\right )^2, \]

and

\[ Y_m(x)=\sqrt {\frac {2}{L_2}}\sin \left (\frac {m\pi y}{L_2}\right ),\qquad \lambda ^Y_m=-\left (\frac {m\pi }{L_2}\right )^2. \]

Solving \(T_{km}''=\lambda _{km} T_{km}\) gives

\[ T_{km}(t)=C_{km}\sin \left (\sqrt {-\lambda _{km}}t\right )+D_{km}\cos \left (\sqrt {-\lambda _{km}}t\right ), \]

so that we obtain the solution

\[ u(t,x,y)=\sum _{k=1}^\infty \sum _{m=1}^\infty \left [C_{km}\sin \left (\sqrt {-\lambda _{km}}t\right )+D_{km}\cos \left (\sqrt {-\lambda _{km}}t\right )\right ]X_k(x)Y_m(y). \]

The initial conditions then become

\[ u^0(x,y)=\sum _{k=1}^\infty \sum _{m=1}^\infty D_{km}X_k(x)Y_m(y),\qquad u^1(x,y)=\sum _{k=1}^\infty \sum _{m=1}^\infty C_{km}\sqrt {-\lambda _{km}}X_k(x)Y_m(y). \]

Using orthonormality of \(X_kY_m\) then gives

\begin{gather*} D_{km}=\int _{[0,L_1]\times [0,L_2]}u^0(x,y)X_k(x)Y_m(y)\,dxdy,\\ C_{km}=\frac {1}{\sqrt {-\lambda _{km}}}\int _{[0,L_1]\times [0,L_2]}u^1(x,y)X_k(x)Y_m(y)\,dxdy. \end{gather*} □