# Heat equation, Fourier cosine transform

by fluidistic
Tags: cosine, equation, fourier, heat, transform
 PF Gold P: 3,188 1. The problem statement, all variables and given/known data 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 $-a  Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,690 That should be ##\nabla^2 u##. You need to find the general solution first. PF Gold P: 3,188  Quote by vela That should be ##\nabla^2 u##. You need to find the general solution first. Oops right, true. Do you mean I should solve the heat equation with say for example separation of variables? If I get the general solution, then why would I need to perform a cosine tranform?  Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,690 Heat equation, Fourier cosine transform Sorry, you're right. You want to set ##\partial u/\partial t=0## and then Fourier transform the equation.  PF Gold P: 3,188 Ok I've checked another source for the definition of the cosine transform, I'll use it instead. I'm having a doubt however. u depends on x,y and t. The cosine transform with respect to y: [itex]\mathbb{F_c}(u)=U_c (p,t)=\int _0 ^{\infty} u \cos (py)dy$. 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?
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PF Gold
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 Quote by 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.
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
$$\mathbb{F_c}\left[\frac{\partial^2 u(x, y)}{\partial x^2}\right] = \frac{\partial^2}{\partial x^2} \mathbb{F_c}[u(x,y)] = \frac{\partial^2}{\partial x^2} u(x,p)$$
 PF Gold P: 3,188 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.
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 Quote by 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.
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.
 PF Gold P: 3,188 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.
 PF Gold P: 3,188 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.
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 Quote by 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.
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.

 Quote by 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.
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).
PF Gold
P: 3,188
 Quote by vela 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 wich 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.
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PF Gold
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 Quote by fluidistic 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.
PF Gold
P: 3,188
 Quote by vela 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.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,690 I didn't notice you were integrating with respect to p. Your latter result is correct. You're just setting x=0 in $$U_c(x,p) = \int_0^\infty u(x,y)\cos py\,dy = B(p)e^{-px}.$$
 PF Gold P: 3,188 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.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,690 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.
 PF Gold P: 3,188 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.

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