
#1
Feb912, 04:39 PM

PF Gold
P: 3,173

1. The problem statement, all variables and given/known data
Problem 817 from Mathew's and Walker's book: Use a cosine transform with respect to y to find the steadystate temperature distribution in a semiinfinite solid [itex]x>0[/itex] when the temperature on the surface [itex]x=0[/itex] is unity for [itex]a<y<a[/itex] and zero outside this strip. 2. Relevant equations Heat equation: [itex]\frac{\partial u}{ \partial t}+k \nabla u =0[/itex]. Cosine Fourier transform: [itex]f(x)=\frac {1} {\pi} \int _0 ^{\infty } g(y) \cos (xy )dy[/itex]. 3. 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 [itex]\frac{\partial u }{\partial t}=0[/itex] since it's a steady state distribution of temperature? This would make [itex]\nabla u =0 \Rightarrow \frac{\partial u }{\partial x }+\frac{\partial u }{\partial y }=0[/itex] (Laplace equation). Should I apply now the cosine transform? 



#2
Feb912, 05:46 PM

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




#3
Feb912, 06:34 PM

PF Gold
P: 3,173

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? 



#4
Feb912, 07:59 PM

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




#5
Feb912, 10:30 PM

PF Gold
P: 3,173

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[/itex]. 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. [itex]\mathbb{F_c} \left ( \frac{\partial u }{\partial x} \right )=\int _0 ^{\infty} \frac{\partial u }{\partial x } \cos (py)dy[/itex]. In order to solve this integral I know I can use integral by parts but I'm not 100% sure that it's worth [itex]p U_s(p,t)u(0,y,t)[/itex] (where [itex]U_s[/itex] 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 [itex]\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)[/itex]. Is this ok so far? 



#6
Feb1012, 02:55 AM

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$$\mathbb{F_c}\left[\frac{\partial^2 u(x, y)}{\partial x^2}\right] = \frac{\partial^2}{\partial x^2} \mathbb{F_c}[u(x,y)] = \frac{\partial^2}{\partial x^2} u(x,p)$$ 



#7
Feb1012, 01:41 PM

PF Gold
P: 3,173

Ok thank you very much vela.
I have a little problem with the cosine transform of [itex]\frac{\partial ^2 u }{\partial y^2}[/itex]. Is it [itex]p^2 U_c (p,x)\frac{\partial u }{\partial y} \big  _{y=0}[/itex]? If so, I don't know how to evaluate the last term. Edit: [itex]\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)[/itex]. Not sure how to evaluate [itex]u(x,0)[/itex] either here. I only know [itex]u(0,0)[/itex] which is worth 1, but no more than this, on the xaxis. 



#8
Feb1112, 09:50 AM

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#9
Feb1112, 10:52 AM

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



#10
Feb1112, 11:18 PM

PF Gold
P: 3,173

So this gives me [itex]\mathbb{F} _c ^{1} [U_c(p,x)]=u(x,y)=\frac{2}{\pi} \int _0 ^{\infty} U_c (p,x) \cos (py)dp[/itex]. Is this ok?
[itex]U_c(p,x)=Ae^{px}+Be^{px}[/itex]. So that [itex]u(x,y)=\frac{2}{\pi} \int _0 ^{\infty } (Ae^{px}+Be^{px} ) \cos (py)dp[/itex]. 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. 



#11
Feb1212, 10:45 PM

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$$U_c(x, p) = A(p)e^{px} + B(p)e^{px}$$ You want a bounded solution as ##x \to \infty##, so you can toss the first term. 



#12
Feb1212, 11:19 PM

PF Gold
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I think something is wrong here. 



#13
Feb1212, 11:32 PM

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



#14
Feb1312, 10:00 AM

PF Gold
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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 [itex]B(p)=\frac{\sin (pa)}{p}[/itex] so that [itex]U_c(p,x)=\frac{e^{px}\sin (pa)}{p}[/itex]. Now time to take the inverse transform. 



#15
Feb1312, 11:26 AM

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



#16
Feb1312, 12:47 PM

PF Gold
P: 3,173

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. [itex]\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}[/itex]. This would be the answer to the problem but I'm hoping to simplify this result. Not sure how to tackle that integral. 



#17
Feb1312, 01:03 PM

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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. 



#18
Feb1312, 01:58 PM

PF Gold
P: 3,173

Right, now I get [itex]u(x,y)=\frac{1}{\pi} \{ \int_0^{\infty} \frac{\sin [p(y+a)]e^{px}}{p}dp + \int_0^{\infty} \frac{\sin [p(ay)]e^{px}}{p}dp \}[/itex].
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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