I have to find the solution of the problem $$ u_t(t,x)=u_{xx}(t,x)+2e^{t-x} \text{ in set } R_+\times R_+ $$ With boundary conditions: $$ \begin{cases} u(0,x)=7\cos x &\text{ for } x>0\\ u_x(t,0)=-2te^{t-x} &\text { for } t>0 \end{cases} $$

There's also the hint to look for a solutions as $ u(x,t)=v(x,t)+a(t)e^{-x}$. After pluging a hint to equation I solved an ODE for $a(t)$ and got this problem: $$ \begin{cases} v_t=v_{xx} &\text{ in } R_+ \times R_+ \\ v(x,0)=7\cos (x) &x>0\\ v_x(t,0) = 0 &t>0\end{cases} $$

I don't know how to solve such problem, as there is only one Neumann boundary condition for $x=0$. Usually there were two conditions ($x=0$ and $x=L$). I would appreciate any hint how to deal with this.


1 Answer 1


The relevant topic is "heat equation on half-line", which is rather different from "heat equation on a finite interval", and has more to do with "heat equation on real line". Namely, an equation on half-line is reduced to an equation on the entire line by reflection; the lone boundary condition gives a clue to which equation to use.

Namely, Neumann equation calls for even reflection, since an even function has zero derivative at $x=0$. So one solves an equation on the entire line (using the fundamental solution) with the initial data extended so that $v(-x,0)=v(x,0)$. In your case this extension is just $7\cos x$ itself, since cosine is already an even function.

But that's in general. Computing the convolution of the fundamental solution with the cosine is a lot more work than realizing that $(\cos 7x)'' = -49 \cos 7x$, and therefore $e^{-49t}\cos 7x$ works.


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