Hi, @Sam, I am looking for numeric solution. For an approximation, a radius r that is sufficiently large (but not infinity) should be enough. When the coefficient of the initial condition is (i.e., ), the plot of M(r, t) as a function of time t and the distance r from the center of the sphere is displayed below:
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M[r,0]
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usol=Block{ϵ=$MachineEpsilon},NDSolveValueD[M[r,t],t]==-D[M[r,t],r,r]-+,M[r,0],[ϵ,t]==0,M[1000,t]0,M,{r,ϵ,4},{t,0,2},Method{"MethodOfLines","SpatialDiscretization"{"TensorProductGrid","MinPoints"12000}};G1=Plot3D4Norm[usol[r,t]],{r,0,3},{t,0,2},AxesLabelStyle["r ",Italic],Style["t",Italic],Rotate"M(r, t)",
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(D[M[r,t],r])
2D[M[r,t],r]
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When the coefficient of the initial condition is larger, such as , as indicated by the red line in the plot below, the value of M(r, t) increases around that time, causing the graph to be concave at that moment.
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vsol=Block{ϵ=$MachineEpsilon},NDSolveValueD[M[r,t],t]==-D[M[r,t],r,r]-+,M[r,0],[ϵ,t]==0,M[1000,t]0,M,{r,ϵ,4},{t,0,2},Method{"MethodOfLines","SpatialDiscretization"{"TensorProductGrid","MinPoints"21000}};xv=NArgMax[Norm[vsol[0,t]],{t,0.1,0.5}];G2=ShowPlot3D2Norm[vsol[r,t]],{r,0,3},{t,0,2},PlotRange->All,AxesLabelStyle["r ",Italic],Style["t",Italic],Rotate"M(r, t)",,ViewPoint{3,-2.2,4.1},ParametricPlot3D[{r,xv,2Norm[vsol[r,xv]]},{r,0,3},PlotStyleDirective[Red,Thickness0.005]]
-M[r,t]
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(D[M[r,t],r])
2D[M[r,t],r]
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When the coefficient is greater than 0.8, the equation becomes difficult to solve. I am currently searching for a solution to this equation with a coefficient of 1 or a larger value, even if it is only a numerical solution.