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Get the analytic solution of transient heat equation in Mathematica?

Posted 4 months ago
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This is my problem: This is the transient heat transfer problem in 1D.

enter image description here

and this is my code:

I can't get the analytic solution. Anyone could help me? Thank you very much!

2 Replies
Posted 4 months ago

Thank you very much!

Workaround:

First we solve PDE only with first two ic's.

$Version
(*12.1.0 for Microsoft Windows (64-bit) (March 14, 2020)*)

Heat = D[T[y, t], t] - α*D[T[y, t], {y, 2}] == 0;
ic = {T[y, 0] == 0, T[0, t] == 1};
sol = DSolve[{Heat, ic}, T[y, t], {y, t}, Assumptions -> {α > 0, t > 0, y > 0}]

(*{{T[y, t] -> (
   y Inactive[Integrate][
     E^(-(y^2/(4 t α - 4 α K[2])))/(t - K[2])^(
     3/2), {K[2], 0, t}, Assumptions -> True])/(
   2 Sqrt[π] Sqrt[α])}}*)

sol // Activate (* Mathematica Can't find Integral ? *)
(*{{T[y, t] -> (
   y Inactive[Integrate][
     E^(-(y^2/(4 t α - 4 α K[2])))/(t - K[2])^(
     3/2), {K[2], 0, t}, Assumptions -> True])/(
   2 Sqrt[π] Sqrt[α])}}*)

Workaround to solve integral:

  FullSimplify[
    InverseMellinTransform[
      Integrate[
       MellinTransform[(
        y E^(-((A*y^2)/(4 t α - 4 α K[2])))/(t - K[2])^(
         3/2))/(2 Sqrt[π] Sqrt[α]), A, s], {K[2], 0, t}, 
       Assumptions -> {t > 0, α > 0, y > 0}], s, A] /. A -> 1, 
    Assumptions -> {t > 0, y > 0, α > 0}] // Expand
 (*Erfc[y/(2 Sqrt[t α])]*)

We check if this solution fulfills Boundary Condition $T(\infty ,t)=0$

 Limit[Erfc[y/(2 Sqrt[t α])], y -> Infinity, 
  Assumptions -> {α > 0, y > 0, t > 0}]
 (* 0 *) (*Check OK*)

We check if this solution fulfills Initial Condition $T(0 ,t)=1$

Limit[Erfc[y/(2 Sqrt[t α])], y -> 0, Assumptions -> {t > 0, α > 0}]
  (*1 *)

We check if this solution fulfills Initial Condition $T(y ,0)=0$

 Limit[Erfc[y/(2 Sqrt[t α])], t -> 0, Assumptions -> {y > 0, α > 0}, Direction -> -1]
 (* 0 *)

Check if solution is true:

 Heat /. {T -> Function[{y, t}, Erfc[y/(2 Sqrt[t α])]]}
 (*True*)

Then solution is:

$$T(y,t)=\text{erfc}\left(\frac{y}{2 \sqrt{t \alpha }}\right)$$

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