## 4 Numerical Solution to Problem (1), (2) on a Rectangle  
We use the change  
\[x:=x,\ \ y:=\sqrt{\frac{a}{c}}y,\ \ u:=u\]  
to write the problem (4), (5) in the form  
\[\frac{\partial u}{\partial t}=\mathcal{L}u+F,\ \ \mathcal{L}u=a\bigg{(}\frac{ \partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}\bigg{)}=a \triangle u, \tag{18}\]  
\[F=F(x,y,t)=\exp\bigg{(}-\beta t-\mu x-\eta\sqrt{\frac{a}{c}}y\bigg{)}f(t) \delta\bigg{(}x-x_{0},\sqrt{\frac{a}{c}}(y-y_{0})\bigg{)},\]  
\[u(x,y,0)=u_{0}(x,y)=\exp\bigg{(}-\mu x-\eta\sqrt{\frac{a}{c}}\bigg{)}C_{0} \bigg{(}x,\sqrt{\frac{a}{c}}y\bigg{)}. \tag{19}\]  
It is known that the Cauchy problem (18), (19) is solved by [21],  
\[u(x,y,t)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}u_{0}( \alpha,\gamma)G(x,y;\alpha,\gamma,t)\mathrm{d}\alpha\mathrm{d}\gamma\\ +\int_{0}^{t}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}F( \alpha,\gamma,\tau)G(x,y;\alpha,\gamma,t-\tau)\mathrm{d}\alpha\mathrm{d} \gamma\mathrm{d}\tau,\]  
where  
\[G(x,y;\alpha,\gamma,t-\tau)=\frac{1}{4\pi a(t-\tau)}\exp\bigg{[}-\frac{(x- \alpha)^{2}+(y-\gamma)^{2}}{4a(t-\tau)}\bigg{]}\equiv G(x-\alpha,y-\gamma,t- \tau).\]  
Let us take \(\Omega=\{0\leq x\leq X,0\leq y\leq Y\}\). Then, the solution to the initial boundary problem  
\[\frac{\partial v}{\partial t}=\mathcal{L}v+F,\ \ (x,y,t)\in Q_{T}=\Omega\times(0,T),\]  
\[v(x,y,0)=u_{0}(x,y),\ \ v(x,y,t)=0,\ \ (x,y)\in\partial\Omega,\ \ 0<t<T\]  
is given by [21]  
\[v(x,y,t)=\int_{-\infty}^{X}\int_{-\infty}^{Y}u_{0}(\alpha,\gamma )V(x,y;\alpha,\gamma,t)\mathrm{d}\alpha\mathrm{d}\gamma\\ +\int_{0}^{t}\int_{-\infty}^{X}\int_{-\infty}^{Y}F(\alpha,\gamma, \tau)V(x,y;\alpha,\gamma,t-\tau)\mathrm{d}\alpha\mathrm{d}\gamma\mathrm{d} \tau,\]  
where  
\[V(x,y;\alpha,\gamma,t)=\frac{4}{XY}\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\sin \frac{n\pi x}{X}\sin\frac{n\pi\alpha}{X}\sin\frac{m\pi y}{Y}\sin\frac{m\pi \gamma}{Y}\exp\bigg{[}-\pi^{2}\bigg{(}\frac{n^{2}}{X^{2}}+\frac{m^{2}}{Y^{2}} \bigg{)}at\bigg{]}.\]  
Now, we can find Dirichlet boundary conditions for the solution of problem (18), (19) from formula (17), which now takes the form\[u(x,y,t)\Big{|}_{\partial\Omega}=\int_{0}^{t}\Bigg{(}\int_{0}^{X} \underbrace{G(x,y,t;\alpha,0,\tau)}_{G(x-\alpha,y,t-\tau)}\frac{\partial v}{ \partial y}(\alpha,0,\tau)\mathrm{d}\alpha\\ -\int_{0}^{Y}\underbrace{G(x,y,t;X,\gamma,\tau)}_{G(x-X,y-\gamma, t-\tau)}\frac{\partial v}{\partial x}(X,\gamma,\tau)\mathrm{d}\gamma+\int_{0}^{X}G(x- \alpha,y-Y,t-\tau)\frac{\partial v}{\partial y}(\alpha,Y,\tau)\mathrm{d} \alpha\\ -\int_{0}^{X}G(x,y-\gamma,t-\tau)\frac{\partial v}{\partial x}(0, \gamma,\tau)\mathrm{d}\gamma\Bigg{)}\mathrm{d}\tau.\]  
Once the boundary conditions are derived, the problem (4), (5), which is imposed on an infinite domain, is transformed to a problem posed on a finite domain. This action has several advantages. First of all, now the computations become feasible and even efficient, since the domain could be arbitrarily small. Secondly, the transformation is exact, i.e., the transformed problem is equivalent to the original one on the bounded domain, which means that there is no truncation error. This is in contrast to the classical methods, which require the truncated domain to be large, thus, introducing both truncation error and heavy computational load.