Heat equation, Fourier cosine transform

[fluidistic]

Homework Statement

Problem 8-17 from Mathew's and Walker's book:
Use a cosine transform with respect to y to find the steady-state temperature distribution in a semi-infinite solid x>0 when the temperature on the surface x=0 is unity for -a<y<a and zero outside this strip.

Homework Equations

Heat equation: \frac{\partial u}{ \partial t}+k \nabla u =0.
Cosine Fourier transform: f(x)=\frac {1} {\pi} \int _0 ^{\infty } g(y) \cos (xy )dy.

The Attempt at a Solution

I've made a sketch of the situation, I don't think I have any problem figuring out the situation.
Now I'm stuck. Should I perform "brainlessly" a cosine transform to the heat equation as it is, or should I set \frac{\partial u }{\partial t}=0 since it's a steady state distribution of temperature? This would make \nabla u =0 \Rightarrow \frac{\partial u }{\partial x }+\frac{\partial u }{\partial y }=0 (Laplace equation). Should I apply now the cosine transform?

 

[vela]

That should be $\nabla^2 u$. You need to find the general solution first.

 

[fluidistic]

That should be $\nabla^2 u$. You need to find the general solution first.

Oops right, true.
Do you mean I should solve the heat equation with say for example separation of variables? If I get the general solution, then why would I need to perform a cosine tranform?

 

[vela]

Sorry, you're right. You want to set $\partial u/\partial t=0$ and then Fourier transform the equation.

 

[fluidistic]

Ok I've checked another source for the definition of the cosine transform, I'll use it instead. I'm having a doubt however.
u depends on x,y and t.
The cosine transform with respect to y: \mathbb{F_c}(u)=U_c (p,t)=\int _0 ^{\infty} u \cos (py)dy. I have no problem with this. I notice that in my exercise the range of y is from negative to positive infinity rather than 0 to positive infinity; but it doesn't matter, I can solve it from 0 to infinity and then use the fact that the function u is symmetric with respect to the x axis, I believe.
\mathbb{F_c} \left ( \frac{\partial u }{\partial x} \right )=\int _0 ^{\infty} \frac{\partial u }{\partial x } \cos (py)dy. In order to solve this integral I know I can use integral by parts but I'm not 100% sure that it's worth p U_s(p,t)-u(0,y,t) (where U_s is the sine transform) because the derivative of u is with respect to x while the integration is with respect to y. This would also mean that \mathbb{F_c} \left ( \frac{\partial ^2 u }{\partial x^2} \right )=-p^2U_c (p,t)-\frac{\partial u }{\partial x}(0,y,t).
Is this ok so far?

 

[vela]

Ok I've checked another source for the definition of the cosine transform, I'll use it instead. I'm having a doubt however.
u depends on x,y and t.

It's actually only a function of x and y because you're looking for the steady-state solution.

The cosine transform with respect to y: \mathbb{F_c}(u)=U_c (p,t)=\int _0 ^{\infty} u \cos (py)dy. I have no problem with this. I notice that in my exercise the range of y is from negative to positive infinity rather than 0 to positive infinity; but it doesn't matter, I can solve it from 0 to infinity and then use the fact that the function u is symmetric with respect to the x axis, I believe.

That's right. You can use the cosine transform because the boundary condition is an even function of y.

\mathbb{F_c} \left ( \frac{\partial u }{\partial x} \right )=\int _0 ^{\infty} \frac{\partial u }{\partial x } \cos (py)dy. In order to solve this integral I know I can use integral by parts but I'm not 100% sure that it's worth p U_s(p,t)-u(0,y,t) (where U_s is the sine transform) because the derivative of u is with respect to x while the integration is with respect to y. This would also mean that \mathbb{F_c} \left ( \frac{\partial ^2 u }{\partial x^2} \right )=-p^2U_c (p,t)-\frac{\partial u }{\partial x}(0,y,t).
Is this ok so far?

You can't integrate by parts like that because the derivative is with respect to x, but the integration is with respect to y. What you can do is switch the order of integration and differentiation, so you'll end up with
$$\mathbb{F_c}\left[\frac{\partial^2 u(x, y)}{\partial x^2}\right] = \frac{\partial^2}{\partial x^2} \mathbb{F_c}[u(x,y)] = \frac{\partial^2}{\partial x^2} u(x,p)$$

 

[fluidistic]

Ok thank you very much vela.
I have a little problem with the cosine transform of \frac{\partial ^2 u }{\partial y^2}. Is it -p^2 U_c (p,x)-\frac{\partial u }{\partial y} \big | _{y=0}? If so, I don't know how to evaluate the last term.
Edit: \mathbb{F_c} \left ( \frac{\partial u (x,y)}{\partial y} \right )=[u(x,y)\cos (py)]^{y=\infty}_{y=0}+p\int _0^{\infty } u(x,y)\sin (py)dy=-u(x,0)+p U_s (p,x). Not sure how to evaluate u(x,0) either here. I only know u(0,0) which is worth 1, but no more than this, on the x-axis.

 

[vela]

Ok thank you very much vela.
I have a little problem with the cosine transform of \frac{\partial ^2 u }{\partial y^2}. Is it -p^2 U_c (p,x)-\frac{\partial u }{\partial y} \big | _{y=0}? If so, I don't know how to evaluate the last term.

Yes, it is. From the symmetry of the physical problem, we know u(x,y) will be symmetric about the x-axis. What does this imply about the derivative at y=0?

Edit: \mathbb{F_c} \left ( \frac{\partial u (x,y)}{\partial y} \right )=[u(x,y)\cos (py)]^{y=\infty}_{y=0}+p\int _0^{\infty } u(x,y)\sin (py)dy=-u(x,0)+p U_s (p,x). Not sure how to evaluate u(x,0) either here. I only know u(0,0) which is worth 1, but no more than this, on the x-axis.

This shouldn't matter because $\frac{\partial u (x,y)}{\partial y}$ isn't in the problem.

 

[fluidistic]

Ok thank you vela!
This means that \frac{\partial u }{\partial y} \big | _{y=0}=0.
Thus the PDE is equivalent to \frac{\partial ^2 U_c (p,x)}{\partial x^2}-p^2 U_c (p,x)=0. Since p>0, U_c(p,x)=Ae^{px}+Be^{-px}. Now I think it's time to take the inverse cosine transform.

 

[fluidistic]

So this gives me \mathbb{F} _c ^{-1} [U_c(p,x)]=u(x,y)=\frac{2}{\pi} \int _0 ^{\infty} U_c (p,x) \cos (py)dp. Is this ok?
U_c(p,x)=Ae^{px}+Be^{-px}. So that u(x,y)=\frac{2}{\pi} \int _0 ^{\infty } (Ae^{px}+Be^{-px} ) \cos (py)dp. This doesn't look a correct answer to me though, let alone how to simplify it and calculate A and B from the boundary conditions.

 

[vela]

Ok thank you vela!
This means that \frac{\partial u }{\partial y} \big | _{y=0}=0.
Thus the PDE is equivalent to \frac{\partial ^2 U_c (p,x)}{\partial x^2}-p^2 U_c (p,x)=0. Since p>0, U_c(p,x)=Ae^{px}+Be^{-px}. Now I think it's time to take the inverse cosine transform.

Remember that the "constants" can still depend on p. That is,
$$U_c(x, p) = A(p)e^{px} + B(p)e^{-px}$$ You want a bounded solution as $x \to \infty$, so you can toss the first term.

So this gives me \mathbb{F} _c ^{-1} [U_c(p,x)]=u(x,y)=\frac{2}{\pi} \int _0 ^{\infty} U_c (p,x) \cos (py)dp. Is this ok?
U_c(p,x)=Ae^{px}+Be^{-px}. So that u(x,y)=\frac{2}{\pi} \int _0 ^{\infty } (Ae^{px}+Be^{-px} ) \cos (py)dp. This doesn't look a correct answer to me though, let alone how to simplify it and calculate A and B from the boundary conditions.

Before you take the inverse transform, you want to incorporate the boundary condition for x=0 by doing essentially what was done on pages 242 and 243 in Mathews and Walker to determine B(p).

 

[fluidistic]

Remember that the "constants" can still depend on p. That is,
$$U_c(x, p) = A(p)e^{px} + B(p)e^{-px}$$ You want a bounded solution as $x \to \infty$, so you can toss the first term.

I am a bit confused here. I want u(x,y) to be bounded when x tends to infinity. I guess you mean that this also imply that A(p) must be worth 0 in which case it's something I have to digest.

Before you take the inverse transform, you want to incorporate the boundary condition for x=0 by doing essentially what was done on pages 242 and 243 in Mathews and Walker to determine B(p).

I get U_c(p,0)=B(p)=\int _0^{\infty} u(0,y) \cos (py)dp.
I think something is wrong here.

 

[vela]

I get U_c(p,0)=B(p)=\int _0^{\infty} u(0,y) \cos (py)dp.
I think something is wrong here.

Why? That's correct. You were given what u(0,y) is equal to.

EDIT: Oops, missed that you were integrating with respect to p.

 

[fluidistic]

Why? That's correct. You were given what u(0,y) is equal to.

True but it's not single valued. It depends on y actually so this makes B depend on y too. Furthermore for -a<y<a, B(p)=\int _0^{\infty } \cos (py ) dp which isn't definied.
I think that if the integration was with respect to y rather than p, I would have less problems.
If I integrate with respect to y rather than p, I get B(p)=\frac{\sin (pa)}{p} so that U_c(p,x)=\frac{e^{-px}\sin (pa)}{p}. Now time to take the inverse transform.

 

[vela]

I didn't notice you were integrating with respect to p. Your latter result is correct. You're just setting x=0 in
$$U_c(x,p) = \int_0^\infty u(x,y)\cos py\,dy = B(p)e^{-px}.$$

 

[fluidistic]

No problem vela, so far you've been of so much help for me...
I'm stuck at solving the integral when taking the inverse transform. \mathbb{F_c}^{-1} [U_c (p,x)]=u(x,y)=\frac{2}{\pi} \int _0^{\infty} \frac{e^{-px}\sin (pa) \cos (py) dp}{p}. This would be the answer to the problem but I'm hoping to simplify this result. Not sure how to tackle that integral.

 

[vela]

Use a trig identity on $2\sin (pa)\cos (py)$. You'll end up with two integrals of the form
$$I=\int_0^\infty \frac{\sin kp}{p} e^{-px}\,dp,$$ where k is a constant, which is the Laplace transform of (sin kp)/p.

 

[fluidistic]

Right, now I get u(x,y)=\frac{1}{\pi} \{ \int_0^{\infty} \frac{\sin [p(y+a)]e^{-px}}{p}dp + \int_0^{\infty} \frac{\sin [p(a-y)]e^{-px}}{p}dp \}.
Now I have to use the residue theorem to calculate both integrals?

Edit: Hmm probably not... If I change p by z, the integral has no residue in z=0 which probably means it has no pole in z=0? Strange.

 

[vela]

Hint:
$$\int_0^k \cos kp\,dk = \frac{\sin kp}{p}$$

 

[fluidistic]

Hint:
$$\int_0^x \cos kx\,dx = \frac{\sin kx}{k}$$

Hmm, definitely not integration by parts. I don't really know right now.

 

[vela]

I edited my hint to make it more suggestive. Use it to evaluate the integral in post #17.

 

[fluidistic]

Ok vela, I've been helped on that particular integral (though I still have doubts) by JJacquelin.
I think the main idea was to consider the integrand as a function of k, then differentiate it with respect to k. Then integrate this result (from 0 to infinity) with respect to p (it's slightly less "ugly" than the original integral, though I'm still unsure how to perform that integral but I will ask for help). Then integrate this result with respect to k and take the constant of integration equal to 0. All in all this gives \int _0 ^{\infty }\frac{\sin (kp)e^{-px}}{p}dp=\arctan \left ( \frac{k}{x} \right ).
Hence u(x,y)=\frac{1}{\pi} \arctan \left ( \frac{y+a}{x} \right )+\frac{1}{\pi} \arctan \left ( \frac{a-y}{x} \right ).
I'd like to know whether your method is the same. If it's a different, I'd like to try it but I don't see how I can use your hint so far.

Edit: I just see that you just replied to my last post. I'll check this out.

 

[vela]

It's slightly different but yields the same result.

 

[fluidistic]

I=\int _0 ^{\infty } \int _0^{k} \cos (kp)dke^{-xp}dp=\int _0^k \int _0 ^{\infty } \cos (kp)e^{-px}dpdk. I'm stuck on that integral with respect to p (integral by parts failed and I don't see any useful substitution so I used Wolfram Alpha for it). I=\int _0 ^k \frac{e^{-xp}[k \sin (kp)-x \cos (kp)]}{k^2+x^2} \big | _0 ^{\infty } dk =\int _0 ^k \frac{x}{k^2+x^2}dk=\arctan \left ( \frac{k}{x} \right ).

Edit: Apparently there are at least 2 ways to tackle down that integral (the one I used wolfram alpha). A double integration by parts or a rewriting of the cos term in function of exponentials. I'll be looking forward this.
Thank you very much for all vela. So the answer is post #22 is correct, right?

 

[vela]

I=\int _0 ^{\infty } \int _0^{k} \cos (kp)dke^{-xp}dp=\int _0^k \int _0 ^{\infty } \cos (kp)e^{-px}dpdk. I'm stuck on that integral with respect to p (integral by parts failed and I don't see any useful substitution so I used Wolfram Alpha for it).

As you noted, you have to integrate by parts twice. Or just look it up in a Laplace transform table, since that's what the integral is.

Thank you very much for all vela. So the answer is post #22 is correct, right?

Yes, it matches what I found. You can verify that at x=0, it reproduces the boundary condition and as x→∞, the solution goes to 0, as you'd expect. It doesn't do anything crazy, so it looks like a valid solution.

 

[fluidistic]

Instead of starting a new thread about sine/cosine Fourier transform I'll ask here.
Basically I wonder why the problem statement made it clear to take a cosine Fourier transform instead of just a Fourier transform. Is it because it is obvious that the solution to the PDE, u(x,y) is an even function for any fixed x? So that a common Fourier transform would lead to a cosine Fourier transform anyway.
P.S.: Just for fun, I think I've plotted the solution function (see post 9 and maybe 10 of https://www.physicsforums.com/showthread.php?t=577993).

 

[vela]

It may have been simply for pedagogical reasons.

 

[fluidistic]

It may have been simply for pedagogical reasons.

Hmm ok I see. The point of my question was more like "taking the Fourier transform is the same as taking the cosine Fourier transform when the function which we wish to take the transform is even". Is what is inside " " right?

 

[vela]

Yes. That's where the Fourier cosine transform comes from, right? You plug the function into the regular Fourier transform integral and then simplify it using the fact the function is even.

 

[fluidistic]

Yes. That's where the Fourier cosine transform comes from, right? You plug the function into the regular Fourier transform integral and then simplify it using the fact the function is even.

Thank you, I wasn't 100% sure. That's exactly what I wanted to know.