Checking Boundary Conditions for Candidate Solutions to PDE

Consider the one-dimensional heat equation $u_{xx}=u_t$ with the boundary data

$$u(x, 0)=f(x), \quad u(0, t)=u(L, t)=0.$$

The standard method of solving this equation is by finding a candidate family of solutions, by means of separation of variables, and then checking that the candidate $u(x, t)$ indeed satisfies both the equation and the boundary data. Regarding showing that the expected solution has the asserted boundary values , Knapp in his Advanced real analysis book writes "This amounts to interchanging infinite sums with passages to the limit as certain variables tend to the boundary, and the following result can often handle that."

Proposition. Let $X$ be a set, let $Y$ be a metric space, let $A_n (x)$ be a sequence of complex-valued functions on $X$ such that $\sum_{n=1}^\infty|A_n (x)|$ converges uniformly, and let $B_n (y)$ be a sequence of complex-valued functions on $Y$ such that $|B_n (y)| \le1$ for all n and y and such that $\lim_{y\to y_0} B_n (y) = B_n (y_0 )$ for all $n$. Then

$$\lim_{y\to y_0}\sum_{n=1}^\infty A_n(x)B_n(y)=\sum_{n=1}^\infty A_n(x)B_n(y_0)$$ and the convergence is uniform in $x$ if, in addition to the above hypotheses, each $A_n (x)$ is bounded.

My question is why do we need to check, for instance, $\lim_{t\to 0}u(x, t)=f(x)$, rather than $u(x, 0)=f(x)$ by simply plugging in $t=0$.

Edit: I suspect that this is because Knapp wants the function $u$ be continuous over its domain of definition. (See the comments below). However Knapp writes "The conditions $u(0, t)=u(L, t)=0$ may be verified in the same way" by the above proposition which literally means that he uses the proposition in evaluation of $u$ at boundary points which to me doesn't sound right.

• Are the function $A_n$ and $B_n$ continous? Do you know that the solution is continuous? The differential equation is usually only assumed to hold inside an open domain, i.e. on $(0,L)\times (0,T)=\Omega_T$ but then the function $$u(x,t)=\begin{cases}1 &(x,t)\in\Omega_T\\ f(x) & t=0\\ 0 &x\in\{0,L\} \end{cases}$$ solves the equation but obviously it is not a "good" solution. Does this help? In particular note that such fourier series may oscillate and thus it may fail that $u$ attains the boundary/initial data Commented Aug 11, 2014 at 6:09
• @Quickbeam2k1 Regarding your first question, the functions $A_n$ and $B_n$ need not be continuous. Indeed, it doesn't make sense to talk about continuity of $A_n$ a priori as $X$ is a set. As for the second question, Knapp, just like many other authors, is not explicit about the continuity of the solutions. My understanding is also that in order to rule out uninteresting solutions such as the ones as you have mentioned, we need to impose extra conditions. But in this case, why don't we write the boundary conditions in the limit form from the beginning?
– EPS
Commented Aug 13, 2014 at 2:47
• have you read the examples in the book? I think the problem could be that $u(x,0)$ is not defined, note the sign $\tilde$ for the fourier series in particular, this can be a problem in the corners (x,t)=(0,0). What if $f$ is not continuous what should $f(x)$ be in a jump point? But I don't see any comments on this there, sorry :) Commented Aug 13, 2014 at 6:46
• $u(x,0)$ is defined in all examples in the book since $u$ is obtained from the method of separation of variables. Sorry, I don't understand your comment about the relation between the discontinuity of $f$ and the requirement for checking $\lim_{t\to 0}u(x,t)=f(x)$.
– EPS
Commented Aug 13, 2014 at 15:46
• @Quickbeam2k1 you should make your comment an answer. As far as I am concerned, your answer is spot on, in the form of rhetorical questions. Commented Aug 14, 2014 at 1:17

Are the functions An and Bn continous? Do you know that the solution is continuous? The differential equation is usually only assumed to hold inside an open domain, i.e. on $(0,L)\times(0,T)=\Omega_T$ but then the function $$u(x,t)=\begin{cases}1& (x,t)\in \Omega_T,\\ f(x)& t=0,\\ 0& x\in\{0,L\}\end{cases},$$ solves the equation but obviously it is not a "good" solution. Does this help? In particular note that such fourier series may oscillate and thus it may fail that u attains the boundary/initial data