Solving a PDE with the following boundary problem with arbitrary constant $b$:

$$u(0,t)=F(t)=b\int_0^\infty u(a,t)\mathrm{d}a$$

Hint given in the question is as follows:

Split this integral in two as it is a non-local boundary condition:

$$F(t)=b\int_0^tu(a,t)\mathrm{d}a+b\int_t^\infty u(a,t)\mathrm{d}a$$

As it turns out, the solution for $u(a,t)$ is as follows;

$$u(a,t)=u_0(a-t)e^{-\frac{1}{2}\mu t^2}$$

Where $\mu $ is an arbitrary constant.

So, the integral is now of the form:

$$u_0(-t)=b\int_0^t u_0(a-t)\mathrm{d}a+b\int_t^\infty u_0(a-t)\mathrm{d}a$$

Should I use Leibniz's integral rule, or is there another integral rule to use here? I'm not sure.

I need to show it is a well-posed problem, so I need to do something with this integral.

  • $\begingroup$ Not sure what you need, but a easy change of variable simplifies the expression a little. $$u_0(-t)=b\int_{-t}^0 u_0(y)\mathrm{d}y+b\int_0^\infty u_0(y)\mathrm{d}y$$ $\endgroup$ – Rafa Budría Mar 17 at 14:42

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