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Consider the heat equation $$u_t = u_{xx}$$ for $t \ge 0$, $0 \le x \le L$, given boundary conditions $$u(0,t) = u(L,t) = f(t)$$ and an initial condition $$u(x,0) = g(x)$$ for some continuous functions $g(x)$ on $[0,T]$ and $f(t)$ on $[0,\infty)$.

Is there an explicit solution for $u$? In particular, I was wondering if $u$ can be expressed in terms of some integral involving $g$ and $f$.

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Imposing some conditions on $f$ and $g$ the solution can be represented via Green's function G: $$ u(x,t)=\int_0^L G(x,y,t)g(y)\,dy+\int_0^t\partial_y G(x,0,t-\tau)f(\tau)\,d\tau- $$ $$ \int_0^t\partial_y G(x,L,t-\tau)f(\tau)\,d\tau. $$ Green's function for the first BVP on a segment can be written out explicitly as series, see ch. 3, $\S7$ in A. Friedman, Partial differential equations of parabolic type.

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In general, there isn't a solution at all, let alone an explicit one. For example, take $f(t)=0$ and $g(x)=(x-L/2)^2 -L^2/4$. Then, at $t=0$, we have $u_t = u_{xx} =2$ at all $x$, including $x=0$ and $x=L$. This, however, contradicts $u_t =0$ for $x=0$ and $x=L$ as determined by the given $f(t)=0$.

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  • $\begingroup$ as a general comment it might be helpful to try and change of variable like $v(x,t)=u(x,t)- f(t)$ and write out the equation for $v$. At least this way one gets a problem with zero Dirichlet boundary conditions; which I assume should help the intuition $\endgroup$ – Math604 Jun 3 at 14:17
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    $\begingroup$ How about a weak solution, not a classical one? $\endgroup$ – user64494 Jun 3 at 14:32
  • $\begingroup$ ya i would assume the change of variables is fine for a weak solution. The main point was that at least this makes the problem much close to what (I assume) most people are used to considering. $\endgroup$ – Math604 Jun 3 at 15:07
  • $\begingroup$ @user64494 - well, the OP asked for the solution on the closed domain. Be careful what you wish for ... indeed, a weak solution works. $\endgroup$ – Michael Engelhardt Jun 3 at 15:48
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Maple 2019.1 finds its weak solution by

pdsolve({diff(u(x, t), t) = diff(u(x, t), x, x), u(0, t) = f(t), u(L, t) = f(t), u(x, 0) = g(x)}, u(x, t));

, producing $$ u \left( x,t \right) =\sum _{n=1}^{\infty } \left( -2\,{\frac {1}{L} \sin \left( {\frac {\pi\,nx}{L}} \right) {{\rm e}^{-{\frac {{\pi}^{2}{ n}^{2}t}{{L}^{2}}}}}\int_{0}^{L}\! \left( -g \left( \tau \right) +f \left( 0 \right) \right) \sin \left( {\frac {\pi\,n\tau}{L}} \right) \,{\rm d}\tau} \right) +\\\int_{0}^{t}\!\sum _{n=1}^{\infty }2 \,{\frac { \left( {\frac {\rm d}{{\rm d}\tau}}f \left( \tau \right) \right) \left( \left( -1 \right) ^{n}-1 \right) }{\pi\,n}\sin \left( {\frac {\pi\,nx}{L}} \right) {{\rm e}^{-{\frac {{\pi}^{2}{n}^{ 2} \left( t-\tau \right) }{{L}^{2}}}}}}\,{\rm d}\tau+f \left( t \right) $$ In particular,

pdsolve({diff(u(x, t), t) = diff(u(x, t), x, x), u(0, t) = 0, u(1, t) = 0, u(x, 0) = (x - 1/2)^2 - 1/4}, u(x, t));

$$u \left( x,t \right) =\sum _{n=1}^{\infty }4\,{\frac {\sin \left( n\pi \,x \right) {{\rm e}^{-{\pi}^{2}{n}^{2}t}} \left( \left( -1 \right) ^ {n}-1 \right) }{{n}^{3}{\pi}^{3}}} $$ and

plot3d(Sum(4*sin(n*Pi*x)*exp(-Pi^2*n^2*t)*((-1)^n - 1)/(n^3*Pi^3), n = 1 .. infinity), x = 0 .. 1, t = 0 .. 2, grid = [60, 60]);

enter image description here

Addition. Mathematica produces the same answer by

DSolve[{D[u[x, t], t] == D[u[x, t], {x, 2}], u[0, t] == 0,  

u[1, t] == 0, u[x, 0] == (x - 1/2)^2 - 1/4}, u[x, t], {x, t}]

$$u(x,t)\to \underset{K[1]=1}{\overset{\infty }{\sum }}\frac{4 \left(-1+(-1)^{K[1]}\right) e^{-\pi ^2 t K[1]^2} \sin (\pi x K[1])}{\pi ^3 K[1]^3} $$

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