A really difficult heat-diffusion problem

Question

A very long beam with a quadratic cross-section $0\le x\le L$, $0\le y\le L$ lies on a plane with a temperature of 0 degrees. The other surfaces of the beam are in contact with air which holds 10 degrees temperature. Determine the stationary temperature distribution in the beam, assuming it is of infinite length $-\infty\le z\le \infty$.

The solution:

I am sure that there are two ways to solve this, one using the heat equation, and one using the Laplace equation.

The heat equation requires a time-dimension, which is not given, however they term "stationary" says it all. $\frac{\partial}{\partial t}=0$. So the heat equation for this should be:

$$\nabla^2 u=0$$

which in fact becomes a Laplace equation, where $u=u(x,y,z)$.

There is a problem with ansatz, it is not easy to make an ansatz here, but I shall try. Consider that the x and y dimensions at the origin are both in contact with a zero-temperature surface and a 10-degree temperature air, then we could consider both to be non-zero at the origin, if we use Kelvin degrees. So we assume both are cosinoid, and therefore have the ansatz:

$$u(x,y,z)=\cos\frac{n\pi}{L}x\cos\frac{m\pi}{L}y\cdot u(z)$$

The operator is then:

$$\nabla^2\bigg(\cos\frac{n\pi}{L}x\cos\frac{m\pi}{L}y\cdot u(z)\bigg):=-\frac{\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\big(\pi^2(m^2+n^2)\big)}{L^2}u(z) -u_{zz} $$ This gives

Insert in the PDE:

$$u_{zz} +\frac{\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}(\pi^2(m^2+n^2)\big)}{L^2}u(z) =0$$

Let $$f(x,y)=\frac{\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\big(\pi^2(m^2+n^2)\big)}{L^2}$$, then we have the "solution":

$$u(z)=C_1\cos (f(x,y)\cdot z)+C_2\sin(f(x,y)\cdot z)$$

Insert in the original ansatz and expand:

$$u(x,y,z)=\cos\frac{n\pi}{L}x\cos\frac{m\pi}{L}y\cdot \big(C_1\cos \bigg(\frac{\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\big(\pi^2(m^2+n^2)\big)}{L^2}\cdot z\bigg)+C_2\sin\bigg(\frac{\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\big(\pi^2(m^2+n^2)\big)}{L^2}\cdot z\bigg)$$

Looking only at a level curve of $u(x,y,z)$, with $z=0$, it looks like this

But the function $u(x,y,z)$ is weird, because each time a level curve is made with x or y dimension equal to zero, then the z-term becomes 1.

In fact, considering a similar ODE problem:

$$y''+\cos xy'=0$$ gives a rather exotic solution, so solving the given PDE must be impossible, analytically.

Does anyone have an idea on how to solve this problem?

Thanks

Answer 1

The BVP is well posed and the solution may be represented as an infinite sum. The mistake is in the ansatz $u(x,y)=u(y)\sin(n\pi x/L)$, because the BCs at $x=0$ and $x=L$ are non-homogeneous.

Let $u(x,y)=v(x,y)+C$, with $C=10$. Now $v$ also satisfies Laplace's equation in the square but with the simpler BCs

$$\tag{1} v(x,0)=-C\qquad,\qquad v(x,L)=v(0,y)=v(L,y)=0 $$

Solving (1) by separation of variables $v(x,y)=X(x)Y(y)$ is straightforward see eg here . The result is

$$\tag{2} v(x,y)=\sum\limits_{n=1}^\infty c_n\sin(n\pi x/L)\sinh(n\pi-n\pi y/L)\\ c_n=\frac{2}{L\sinh(n\pi)}\int\limits_0^L dx \ \sin(n\pi x/L)v(x,0) $$

With $v(x,0)=-C$ we find

$$\tag{3} c_n=-\frac{2C}{n\pi \sinh(n\pi)}\left(1-(-1)^n\right) $$

Inserting (3) into (2) we have

$$\tag{4} v(x,y)=-4C\sum\limits_{n=1,3,5\cdots}\frac{\sinh(n\pi-n\pi y/L)}{n\pi\sinh(n\pi)}\sin(n\pi x/L) $$

Before plotting, it's useful to switch to dimensionless variables. Define $\chi=x/L$, $\eta=y/L$, $\phi=u/C$. The BVP is now posed on $0<\chi<1$, $0<\eta<1$, with BCs $\phi=1$ and $\phi=0$. The solution is

$$\tag{5} \phi(\chi,\eta)=1-\sum\limits_{n=1,3,5,\cdots} \frac{4 \sinh(n\pi-n\pi \eta)}{n\pi \sinh(n\pi)}\sin(n\pi \chi) $$

Here is a plot of the $n=1$ term (left) and the partial sum to $11$ terms (right)

Answer 2

By Sals suggestion, I add the proposed solution.

Let the boundaries of the heat-diffusion problem be defined as

We ignore the z-dimension.

The PDE is given therefore by:

$$\nabla^2 u=0 \ \ \ \ \ \ 0\le x\le L, \ \ \ \ 0\le y\le L\\ u(0,y)=10,\ \ \ \ u(L,y)=10 \ \ \ \ \ (1)\\ u(x,0)=0, \ \ \ \ u(x,L)=10 \ \ \ \ \ \ (2)$$

The IC in (1) are Dirichlet homogeneous on x, thus a suitable ansatz is $u(x,y)=\sin\frac{n\pi x}{L}u(y)$

We insert in the PDE and get:

$$-\bigg(\frac{n\pi}{L}\bigg)^2\sin\frac{n\pi x}{L}u(y)+\sin\frac{n\pi x}{L}u_{yy}=0$$

which gives the ODE:

$$u(y)_{yy}-\bigg(\frac{n\pi}{L}\bigg)^2u(y)=0$$

This gives the solution $$u(y)=C_1\cosh\frac{n\pi}{L}y+C_2\sinh\frac{n\pi}{L}y$$

Use I.C. from (2) and obtain:

$$u(y)=\frac{10}{\sinh n\pi}\sinh\frac{n\pi}{L}y$$

Then, the final solution is:

$$u(x,y)=\frac{10}{\sinh n\pi}\sinh\frac{n\pi}{L}y\sin\frac{n\pi}{L}x$$

Which has the plots

Any comment or eventual improvements are welcome!