How to obtain the exact solution of a partial differential equation?

I know that Mathematica can solve a PDE numerically, but I wonder if it is possible to obtain the exact solution. For example, consider the heat equation

$$u_t = \kappa u_{xx}$$

Is it possible to solve it with a set of initial and boundary conditions to calculate the exact equation of $u$

$$u = f(x,t)$$ and $$u(x=0) = f(t)$$

I don't need numeral solution or the graph but the general equations.

EXAMPLE

One dimensional heat flow in an slab, one side is insulated and the other side at a constant flux of heat

$$u(x,0) = U\\ u_x(0,t) = 0\\ u_x(L,t) = T$$

The solution is available from the textbooks. I just wonder, if Mathematica can give us the solution, as we can slightly alter the conditions to find new solutions.

• That depends really on the domain. For example, the fundamental solution for the disk is known and the symbolic solution can be obtained via convolution with it. – Henrik Schumacher Aug 7 '18 at 11:32
• @HenrikSchumacher I am interested in the boundary conditions rather than domain. I would consider one-dimensional. It may seem strange, but I am interested to find the solution by Laplance transformation or so. I add an example. – Kiera Aug 7 '18 at 11:45
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Sometimes you can solve pde's symbollically.

example heat equation:

DSolve[{Derivative[0, 1][u][x, t] == Derivative[2, 0][u][x, t],u[0, t] == f[t]}, u, {x, t}][]

The specific example can be solved with the help of finite Fourier cosine transform and its inversion:

With[{u = u[t, x]},
eq = D[u, t] == κ D[u, x, x];
ic = u == U[x] /. t -> 0;
bc = {D[u, x] == 0 /. x -> 0, D[u, x] == T /. x -> L};]

Format@finiteFourierSinTransform[f_, __] := Subscript[ℱ, s][f]
Format@finiteFourierCosTransform[f_, __] := Subscript[ℱ, c][f]

help[index_] :=
Module[{tset =
finiteFourierCosTransform[{eq, ic}, {x, 0, L}, index] /. Rule @@@ bc /.
HoldPattern@finiteFourierCosTransform[f_ /; ! FreeQ[f, u], __] :> f},
tsol = DSolve[tset, u[t, x], t][[1, 1, -1]]]

tsolgeneral = help[n]

tsolzero = help

tsolfunc[n_] = Piecewise[{{tsolgeneral, n != 0}}, tsolzero]

sol = inverseFiniteFourierCosTransform[tsolfunc[n], n, {x, 0, L}] // transformToIntegrate Let's check the solution numerically with $U=x(1-x),\ L=1,\ κ = 1,\ T = 1$:

U = (# (1 - #) &); L = 1; κ = 1; T = 1;
nsol = NDSolveValue[{eq, ic, bc}, u, {t, 0, 1/10}, {x, 0, 1},
Method -> {"MethodOfLines",
"DifferentiateBoundaryConditions" -> {True, "ScaleFactor" -> 5000}}];

With[{expr =
Block[{C = 20, HoldForm = Identity,
Sum = Function[{expr, lst}, Total@Table[expr, lst], HoldAll]}, sol]},
Manipulate[Plot[{expr, nsol[t, x]}, {x, 0, 1}, PlotRange -> All], {t, 0, 1/10}]]
Clear[U, L, κ, T] Remark

1. Finite Fourier Cosine transform at $n=0$ is calculated separately here because currently finiteFourierCosTransform cannot handle the singularity at $n=0$ properly.

2. The reason why "DifferentiateBoundaryConditions" option is added is explained in this post.

• don't you get the error, NDSolveValue::ibcinc: Warning: boundary and initial conditions are inconsistent.? – Googlebot Aug 7 '18 at 20:34
• @Googlebot It doesn't matter as long as "DifferentiateBoundaryConditions" option is properly set. For more information you can check the last link in my answer. – xzczd Aug 7 '18 at 23:38