Chapter 3 The heat equation on a rectangle
Consider the initial-boundary value problem for the partial differential equation (this PDE is called the heat equation)
\(\seteqnumber{0}{3.}{0}\)\begin{equation} \label {eq:heat:rec} \partial _t u=\partial _{xx}u+\partial _{yy}u,\quad u(t,0,y)=u(t,L_1,y)=u(t,x,0)=u(t,x,L_2)=0,\quad u(0,x,y)=u^0(x,y), \end{equation}
where the initial condition \(u^0:[0,L_1]\times [0,L_2]\to \mR \) is given and \(u:[0,\infty )\!\!\phantom {]}\times [0,L_1]\times [0,L_2]\to \mR \) is to be found. Here \(\partial _tu\) is notation for the partial derivative of \(u\) with respect \(t\) (which we interpret as time) and \(\partial _{xx}u\) and \(\partial _{yy}u\) are notation for the second partial derivatives of \(u\) with respect to \(x\) respectively \(y\). We view \((x,y)\) as the two-dimensional spatial variable which lives on the rectangle \([0,L_1]\times [0,L_2]\) where the sides lengths \(L_1,L_2>0\) are given.
As in Chapter 1, we first consider only the differential equation and look for a solution \(u\) which is separated, i.e. of the form
\[ u(t,x,y)=T(t)X(x)Y(y). \]
This gives
\[ T'XY=TX''Y+TXY'', \]
which leads to
\[ \frac {T'}{T}=\frac {X''}{X}+\frac {Y''}{Y}. \]
Since the left-hand side only depends on \(t\) and the right-hand side only depends on \((x,y)\), there must exist a constant \(\lambda \) such that
\[ \frac {T'}{T}=\lambda =\frac {X''}{X}+\frac {Y''}{Y}, \]
or written differently
\[ T'=\lambda T,\qquad X''Y+Y''X=\lambda XY. \]
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Remark 3.1. Note that this second equation alternatively arises from assuming a separated solution \(\tilde {u}(x,y)=X(x)Y(y)\) for the eigenvalue problem \(\partial _{xx}\tilde {u}+\partial _{yy}\tilde {u}=\lambda \tilde {u}\) for the spatial differential operator \(\partial _{xx}+\partial _{yy}\).
From \(\lambda =\frac {X''}{X}+\frac {Y''}{Y}\) we obtain
\[ \frac {X''}{X}=\lambda -\frac {Y''}{Y}, \]
which since the left-hand side only depends on \(x\) and the right-hand side only depends on \(y\) shows that there must exist a constant \(\mu \) such that
\[ \frac {X''}{X}=\mu ,\qquad \lambda -\frac {Y''}{Y}=\mu , \]
which we can re-write as
\[ X''=\mu X,\qquad Y''=(\lambda -\mu )Y. \]
Changing constants through \(\lambda ^X:=\mu \) and \(\lambda ^Y:=\lambda -\mu \) (so that \(\lambda =\lambda ^X+\lambda ^Y\)) gives
\[ T'=(\lambda ^X+\lambda ^Y)T,\qquad X''=\lambda ^X X,\qquad Y''=\lambda ^Y Y. \]
From the boundary conditions we obtain
\[ X(0)=X(L_1)=0,\qquad Y(0)=Y(L_2)=0. \]
The ODE boundary value problems which we obtain for \(X\) and \(Y\) are therefore identical (up to notation) to that in Chapter 1. Hence we have
\[ X_k(x)=\sqrt {\frac {2}{L_1}}\sin \left (\frac {k\pi x}{L_1}\right ),\qquad \lambda ^X_k=-\left (\frac {k\pi }{L_1}\right )^2, \]
and
\[ Y_m(x)=\sqrt {\frac {2}{L_2}}\sin \left (\frac {m\pi x}{L_2}\right ),\qquad \lambda ^Y_m=-\left (\frac {m\pi }{L_2}\right )^2, \]
so that
\[ \lambda _{km}=\lambda ^X_k+\lambda ^Y_m=-\left (\frac {k\pi }{L_1}\right )^2-\left (\frac {m\pi }{L_2}\right )^2. \]
We therefore obtain the solutions
\[ u_{km}(t,x,y)=C_{km}\e ^{\lambda _{km}t} X_k(x)Y_m(y). \]
To be able to satisfy the initial conditions we consider the sum:
\[ u(t,x,y)=\sum _{k=1}^\infty \sum _{m=1}^\infty C_{km}\e ^{\lambda _{km}t} X_k(x)Y_m(y). \]
The initial condition then becomes
\[ u^0(x,y)=\sum _{k=1}^\infty \sum _{m=1}^\infty C_{km} X_k(x)Y_m(y). \]
As in Chapter 1, we can solve this for \(C_{km}\) by multiplying with \(X_pY_q\) and integrating over \([0,L_1]\times [0,L_2]\) and using orthonormality of \((X_k)\) and \((Y_m)\) to obtain
\[ C_{pq}=\int _{[0,L_1]\times [0,L_2]} u^0(x,y)X_p(x)Y_q(y)\,dxdy. \]