# Iteration show different values in solving a pde in mathematica

I have a pde given below: $$u_{t}-u_{xx}+\frac{1}{4}u=0,$$ with boundary conditions $$u(0,t)=1,\ \ u(1,t)=\frac{1}{2}e^{\frac{-t}{4}}+e^{\frac{-1}{2}},$$ where $$0\leq x\leq 1$$. If the initial iterate is $$u_{0}=(\frac{1}{2}e^{\frac{-t}{4}}+e^\frac{-1}{2}$$-1)x+1$$, \alpha_{n}=0.9$$. Then I use the following iterate to obtain the numerical solution for the above pde:

$$u_{n+1}(x,t)=u_{n}(x,t)+\alpha_{n}\int_{0}^{x}x(1-s)[(u_{n})_{t}-(u_{n})_{xx}+\frac{1}{4}u_{n}]ds+\alpha_{n}\int_{x}^{1}s(1-x)[(u_{n})_{t}-(u_{n})_{xx}+\frac{1}{4}u_{n}]ds$$.

I have coded the above iteration as follows:

\[Delta] = 10^-20;
Clear[u];
u[0 _]:= u[0] = Function[{x,t},(Exp[-t/4]/2+Exp[-1/2]-1)x+1];
a[n_]:= a[n] = 0.9;
u[n_]:= u[n]=Function[{x,t},Evaluate[Chop[Expand[u[n - 1][x,t]+Integrate[Expand[x(1-s) (D[u[n - 1][x,s], s]-D[u[n-1][x,s],{x,2}]+0.25*u[n-1][x,s])],{s,0,x}]+Integrate[Expand[s(1-x)(D[u[n-1][x,s],s]-D[u[n-1][x,s],{x,2}]+0.25*u[n-1][x,s])], {s,x,1}]], \[Delta]]]]
a1a = Table[u[n][0.2, 0.2], {n, 0, 3}]


When I run my code for three iteraions, i get the values for $$(0.2,0.2)$$ as $$1.01643, 0.901749. 0.892542, 1.00427$$, which are no convergent to the required solution. I have attached the table of a paper, I donot where is the mistake in my code. The exact solution of this pde is $$u(x,t)=\frac{1}{2}xe^{\frac{-t}{4}}+e^{\frac{-x}{2}}$$. I also replace $$u[n-1][x,t]$$ by $$u[n][{x,t}]$$ but this not work. I think there is saomething wrong in my code. This example is the example 1 of the paper https://doi.org/10.1007/s40819-016-0289-x

• Is there an a[n] missing from the u[n] definition? Sep 20, 2023 at 21:29
• With 39.95€ the link is not a bargain . Perhaps you can show the main idea of the iteration? Sep 21, 2023 at 7:09
• Is there a comma missing? u[0 _] Sep 21, 2023 at 7:32
• Your exact solution x/2 Exp[-t/4] + Exp[-x/2] doesn't solve the pde! Sep 21, 2023 at 7:41
• Interestingly, the exact solution in the text $$u(x,t) = \frac{1}{2} x~e^{-\frac{t}{4}} +e^{-\frac{x}{2}}$$ is a solution to $$u_t - u_{xx} + \frac{1}{4}u =0~ ,$$ but not (as @UlrichNeumann already pointed out) $$u_t - u_{xx} - \frac{1}{4}u =0 ~.$$
– ydd
Sep 21, 2023 at 19:45