Plotting the solution of the following equation with integral boundary conditions

Question

This is my equation:

E1 = D[u[x, t], t] + D[u[x, t], x] + u[x, t] == 0;
ic = {u[x, 0] == Sin[\[Pi] x]};
bcc = {u[0, t] == Integrate[u[x, t], {x, 0, Infinity}]};

The solve the above system, I used the following command

sol1 = DSolveValue[{E1, ic, bcc}, u[x, t], x, t]

To Plot the solution, I used the following

Plot[Evaluate[Table[sol1, {t, 0, 12}]], {x, 0, 1}, PlotRange -> All, 
 Filling -> Axis]

This provides me error!! it doesn't work!! can anyone help me, please?

Why are there so many exclamation marks ?? I initially ignored this question as the exclamation marks and the please at the beginning stresses me a bit. Then I got curious and saw that you have another post with the same please format. Please and thank you can be included in the body of the text but in my opinion the title should just be informative. I also think a thank you after receiving an answer is better than a please during a question here. The exclamation marks are unnecessary and in my opinion goes against the incentive to help which is probably what you would like. — userrandrand, Commented Oct 4, 2022 at 19:11
Look at the boundary condition for t==0.: Integrate[ Sin[Pi x], {x,0,Infinity}]. Note that this integral does not exists. — Daniel Huber, Commented Oct 4, 2022 at 20:25

user46831 · Accepted Answer · 2022-10-04 21:26:28Z

1

sol1 = DSolveValue[Join[{E1}, ic],  u, {x, 0, \[Infinity]}, {t, 0, \[Infinity]}]

returns

Function[{x, t}, E^-t Sin[\[Pi] (-t + x)]]

To view that you might look at

Plot3D[sol1[x, t], {t, 0, 12}, {x, 0, 1}]

answered Oct 4, 2022 at 21:26

user46831

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$\begingroup$ Thank you for your answer, But, I really can't see where is the boundary conditions? You provide a solution without taking into consideration the boundary conditions!! $\endgroup$
– walid fssm
Commented Oct 5, 2022 at 15:29
$\begingroup$ Yes. I didn't look at it closely, but DSolve produces a general solution and, evidently, that solution is unique given only the initial condition. Notice that there are no unspecified terms in the solution. I suspect that with the boundary condition the problem is over-specified and there may be no solution. $\endgroup$
– user46831
Commented Oct 6, 2022 at 17:13

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