2

If we use the separation of variables to solve $u_t-u_{xx}=-u$, $u=u(x,t)$, then we obtain $f(x)g'(t)-f''(x)g(t)=-f(x)g(t) \iff \frac{g'(t)}{g(t)}+1=\frac{f''(x)}{f(x)}$. The only way that a function of $x$ can equal a function of $t$ is for both functions to be the same constant, say $\lambda$. Thus, we obtain $g'(t)+(1-\lambda)g(t)=0$ and $f''(x)-\lambda f(x)=0$.

This is just a particular case of what we could do by separation of variable. Question : In which case, are we certain that $\lambda \geq 0$?

  • 1
    This depends on the x-variable operator (in this case, the one dimensional Laplacian). The parameter $\lambda$ must be taken nonnegative if and only if the operator has no negative eigenvalues. This is actually a tautology, but helps to introduce the correct language to treat those things: spectral theory. – Giuseppe Negro May 12 '16 at 23:49

0 Answers0