### 4.8 Show the eﬀect of boundary/initial conditions on 1-D heat PDE

The PDE is $\frac {\partial T\left ( x,t\right ) }{\partial t}=k\frac {\partial ^{2}T\left ( x,t\right ) }{\partial x^{2}}$ Problem: given a bar of length $$L$$ , solve the above 1-D heat PDE for 4 diﬀerent boundary/initial condition to show that the solution depends on these.

Mathematica


Clear[y,x,t,k];
SetDirectory[NotebookDirectory[]];
barLength = 4*Pi;
timeDuration = 10;
eq = D[y[x, t], t] == k*D[y[x, t], x, x];
eq = eq /. k -> 0.5;
solveHeat[eq_,bc_,ic_,y_,x_,t_,len_,timeDuration_]:=Module[{sol},
sol=First@NDSolve[{eq,bc,ic},y[x,t],{x,0,len},{t,0,timeDuration}];
Plot3D[y[x,t]/.sol,{x,0,barLength},{t,0,timeDuration},
PlotPoints->30,PlotRange->All,AxesLabel->{"x","time","y[x,t]"},
ImageSize->200,PlotLabel->bc]
];

bc={{y[0,t]==1,y[barLength,t]==1},
{y[0,t]==0,y[barLength,t]==0},
{y[0,t]==1,y[barLength,t]==Exp[-t barLength]},
{y[0,t]==0,y[barLength,t]==0}
};

ic={y[x,0]== Cos[x],
y[x,0]==Sin[x],
y[x,0]==1,
y[x,0]== Cos[x]Sin[x]
};