Consider $$\partial_tu+ \partial_xu-D\partial_{yy}u=0,$$ subject to the initial condition $u(x,y,0) = v(x,y)$ and the boundary conditions at the infinities, $ u(0,0,t) = u(\pm\infty,\pm\infty,t) = 0,$ where $D$ is the diffusion coefficient. To use the Sumudu Transform (ST) method, the ST of the given equation is taken as $$\frac{1}{w}\left(\mathcal{S}[u(x,y,t)]-u(x,y,0)\right)+ \mathcal{S}\left[\partial_xu-D\partial_{yy}u\right]=0,$$ which leads to $$\mathcal{S}[u(x,y,t)]=v(x,y) - w\mathcal{S}\left[\partial_xu-D\partial_{yy}u\right].$$ Applying the inverse ST leads to $$u(x,y,t)=v(x,y) -\mathcal{S}^{-1}\left[ w\mathcal{S}\left[\partial_xu-D\partial_{yy}u\right]\right].$$ Applying the property of ST, which states that $$\mathcal{S}^{-1}\left[ w\mathcal{S}[u(t)]\right]=\int_0^tu(w)dw,$$ then the equation becomes $$u(x,y,t)=v(x,y) -\int_0^t \left[v(x,y)\partial_xu-D\partial_{yy}u\right]dw.$$
How can one complete the solution and plot the graph of the above equation?
Disclaimer: this is not a solution using the Sumudu transform method --- it is a solution using the Fourier transform method.
Let's assume that $u(x,y,t)$ has a Fourier transform $U(k,q,t)$, i.e. $$ u(x,y,t)=\int U(k,q,t)e^{ikx+iqy}\,dk\,dq. \tag{1} $$ Plugging $(1)$ into the PDE $\partial_tu+ \partial_xu-D\partial_{yy}u=0$, we reduce it to the ODE $\partial_tU+(ik+Dq^2)U=0$, whose solution is $$ U(k,q,t)=U(k,q,0)e^{-(ik+Dq^2)t}. \tag{2} $$ Inserting $(2)$ into $(1)$, together with $$ U(k,q,0)=\frac{1}{(2\pi)^2}\int u(x',y',0)e^{-ikx'-iqy'}\,dx'dy', \tag{3} $$ we obtain $$ u(x,y,t)=\frac{1}{(2\pi)^2}\int u(x',y',0)e^{ik(x-x'-t)+iq(y-y')-Dq^2t}dx'dy'dk\,dq. \tag{4} $$ Integration over $k$ and $q$ yields $$ u(x,y,t)=\frac{1}{(2\pi)^2}\int u(x',y',0)\,2\pi\delta(x-x'-t)\, \sqrt{\frac{\pi}{Dt}} e^{-\frac{(y-y')^2}{4Dt}}dx'dy'. \tag{5} $$ Finally, using the initial condition $u(x',y',0)=v(x',y')$ and integrating over $x'$, we arrive at $$ u(x,y,t)=\frac{1}{\sqrt{4\pi Dt}}\int v(x-t,y')e^{-\frac{(y-y')^2}{4Dt}}dy' \qquad(t>0). \tag{6} $$