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I am trying to solve the following Laplace PDE in MMA:

$$\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial z^2}=0$$

subject to the boundary conditions: $$\frac{\partial T}{\partial z}|_{z=0}=0,$$

and, at $z=1$ $$-\frac{\partial T}{\partial z}+A=BT,$$

This is my code:

DSolve[{D[T[x, z], {x, 2}] + D[T[x, z], {z, 2}] == 0, (D[T[x, z], z] /. {z -> 0}) == 0, -(D[T[x, z], z] /. {z -> 1}) + A == B T[x, 1]}, T[x, z], {x, z}]

But it just repeats my equation.

Thank you for kindly help in advance!

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  • $\begingroup$ Can you check to make sure that you don't have any definitions for the symbols you use in your equations? $\endgroup$ Commented Sep 25, 2015 at 5:14
  • $\begingroup$ The syntax in D[T[x, 0],z]is not correct: you should evaluate in $z=0$ after taking the derivative. This yields 0==0 and so your error. Apart from this, are you sure about you boundary/initial conditions? Seems to me you are missing some. $\endgroup$ Commented Sep 25, 2015 at 5:32
  • $\begingroup$ Hi, @anderstood, thanks. Please check my modified post. It seems that MMA has done nothing... $\endgroup$ Commented Sep 25, 2015 at 5:49
  • $\begingroup$ What if you solve only the pde (DSolve[{D[T[x, z], {x, 2}] + D[T[x, z], {z, 2}] == 0}, T[x, z], {x, z}]) and add the boundary conditions afterwards to identify the constants? $\endgroup$ Commented Sep 25, 2015 at 5:55
  • $\begingroup$ Thanks, @anderstood. Your method looks feasible. But I have not realized it by now... $\endgroup$ Commented Sep 25, 2015 at 6:10

2 Answers 2

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Another problem easy to do by separation of variables. However, BC's on x are definitely required to solve the problem. The assumption of exp[i a x] requires BC's that make sinusoidal solutions in x.

pde = D[T[x, z], x, x] + D[T[x, z], z, z] == 0;

With the BC's you give I add 2 more for x. The ones I chose will give sinusoidal solutions. You, of course can choose others

bc1 = T[0, z] == 0; bc2 = T[1, z] == 0; bc3 = (D[T[x, z], z] /. z -> 0) == 0; bc4 = (D[T[x, z], z] /. z -> 1) == A - B T[x, 1];

Separate T in the form

T[x_, z_] = X[x] Z[z]; pde/T[x, z] // Apart (* D[X[x],x,x]/X[x]+D[Z[z],z,z]/Z[z]==0 *)

One terms is dependent on x, the other z, so each must be equal to a constant.

xeq = D[X[x], x, x]/X[x] == -alpha^2 X[x_] = (X[x] /. DSolve[xeq, X[x], x][[1]]) /. {C[1] -> c1, C[2] -> c2}; zeq = D[Z[z], z, z]/Z[z] == alpha^2 Z[z_] = (Z[z] /. DSolve[zeq, Z[z], z][[1]]) /. {C[1] -> c3, C[2] -> c4} (* c3 E^(alpha z)+c4 E^(-alpha z) *)

From the z BC's, it is obvious we want hyberbolics rather than exponentials.

(Z[z] // ExpToTrig) // Collect[#, {Sinh[alpha z], Cosh[alpha z]}] & (* (c3-c4) Sinh[alpha z]+(c3+c4) Cosh[alpha z] *) Z[z_] = % /. {c3 - c4 -> c3, c3 + c4 -> c4}; (* c3 Sinh[alpha z] + c4 Cosh[alpha z] *)

Apply the BC's

bc1

(* c1 (c3 Sinh[\[Alpha] z] + c4 Cosh[\[Alpha] z]) == 0 *)

From which

c1 = 0 T[x, z] (* c2 Sin[alpha x] (c3 Sinh[alpha z] + c4 Cosh[alpha z])*)

Combine constants

c2 = 1; bc2 (* Sin[alpha] (c3 Sinh[alpha z] + c4 Cosh[alpha z]) == 0 *)

From which

alpha = n Pi

with n required to be

$Assumptions = n \[Element] Integers && n > 0; bc3 (* Pi c3 n Sin[Pi n x]==0 *)

From which

c3 = 0; bc4 // Simplify (* c4 Sin[Pi n x] (B Cosh[Pi n]+Pi n Sinh[Pi n])==A *)

Use orthogonality to solve for c4

Integrate[%[[1]] Sin[n Pi x], {x, 0, 1}] == Integrate[%[[2]] Sin[n Pi x], {x, 0, 1}]; c4 = c4 /. Solve[%, c4][[1]] T[x, z] (*-((2 A ((-1)^n - 1) Sin[Pi n x] Cosh[Pi n z])/(Pi n (B Cosh[Pi n] + Pi n Sinh[Pi n]))) *)

The actual T will be an infinite series in n, but obviously from the form of the solution, the even n terms are 0. In order to not compute the 0 terms we can do the following:

Tn[x_, z_] = (T[x, z] /. n -> 2 m - 1) /. m -> n // Simplify (*(4 A Sin[Pi (2 n-1) x] Cosh[Pi (2 n-1) z])/(Pi (2 n-1) (B Cosh[Pi-2 Pi n]+Pi (2 n-1) Sinh[Pi (2 n-1)]))*)

The general solution is the above summed from n = 1 to infinity. To make a plot:

tn = Function[n, #] &@Tn[x, z];

Assign A and B some values.

A = 1; B = 2; Tterms[k_] := Compile[{x, z}, #] &@Total@tn@Range[k] T = Tterms[50]; Plot3D[T[x, z], {x, 0, 1}, {z, 0, 1}, AxesLabel -> {"x", "y", "T"}]

enter image description here

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There is a problem about your solution, it is that DSolve cannot handle it and, therefore, returns you the input unevaluated, as a signal. So, what you can do, if you do not want to solve it analytically (seems a simple task, does not it?) you might want to solve it numerically, that is, using NDSolve. This, of course, will reguire from you to somehow fix the values of A and B, otherwise, however, it can be solved. Try this:

 A = 1; B = 2; << NDSolve`FEM` mesh = ToElementMesh[Rectangle[{0, 0}, {1, 1}]]; nv1 = NeumannValue[A - B*T[x, z], z == 1]; nv2 = NeumannValue[0, z == 0]; dc = DirichletCondition[T[x, z] == 1, x == 0]; sl = NDSolve[{D[T[x, z], {x, 2}] + D[T[x, z], {z, 2}] == nv1 + nv2, dc}, T[x, z], {x, z} \[Element] mesh] (* {{T[x, z] -> InterpolatingFunction[{{0., 1.}, {0., 1.}}, <>][x, z]}} *)

To have a look at the solution evaluate this:

Plot3D[T[x, z] /. sl[[1, 1]], {x, z} \[Element] mesh, ImageSize -> 300]

which should look as follows:

enter image description here

Take into account that I added a Dirichlet condition 

 dc = DirichletCondition[T[x, z] == 1, x == 0];

otherwise there is nothing to look at in the solution, as it was.

Have fun!

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  • $\begingroup$ @ Alexei Boulbitch, Thank you. But I have some troubles in debugging your code. I am using MMA 9. First, what is the meaning of << before your NDSolveFEM``. I guess you want to use Needs? Second, when I run it, MMA gives NDSolve::dvnoarg: "The function T[x,z] appears with no arguments. ". In addition, I think the original two boundary condition are sufficient to solve for the function T(x,z) even analytically. For example, assuming $T(x,z)=\theta(z)\exp[i\alpha x]$, we can obtain $T(x,z)=\frac{A \cosh \alpha z}{B\cosh \alpha + \alpha \sinh \alpha}$. $\endgroup$ Commented Sep 26, 2015 at 3:13
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    $\begingroup$ @ lxy Welcome. To address your questions. 1. << NDSolveFEM` is the short form of Needs... . You may use the latter instead. 2. I am not quite sure, if Mma 9 already has the NDSolveFEM package. If not (which is possible), it is the reason for the strange warnings like "The function T[x,z] appears with no arguments". 3. Indeed the original boundary conditions enable one to solve the equation. The solution, however, was a constant. That's why I added something to see that there is at least some variation, and in this way to check that there is, indeed, a solution. $\endgroup$ Commented Sep 28, 2015 at 7:24
  • $\begingroup$ @ Alexei Boulbitch, Thanks, I confirmed that MMA9 does not have FEM package, and it can run on version 10 :) $\endgroup$ Commented Sep 28, 2015 at 9:07

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