3.6 Applications of Poisson s equation

Transcription

1 William J. Parnell: MT3432. Section 3: Green s functions in 2 and 3D Applications of Poisson s equation The beaut of Green s functions is that the are useful! Let us consider an example here where we determine the stead state heat distribution on a two dimensional domain. In question 5 of Example Sheet 6 ou are asked to consider a three dimensional problem relating to the distribution of electrical charge. This was the problem for which Green originall devised Green s functions! How great is that?! Example 3.4 Suppose that we have a ver large two dimensional domain and suppose that it is so large that boundar effects can be ignored, i.e. the boundar is so far awa that an heat source generates a heat distribution which has decaed to zero along time and distance before the boundar. We use a heat source to heat a circular region C of radius a uniforml with magnitude q and this source is maintained for all times. What is the stead state temperature field that results? You ma use without proof the result { lnβ 2, β 2 <, + 2β cosψ dψ = 3.59 β 2, β 2 >. Since we can ignore boundar effects we can assume that the circular domain C has its centre at the origin of our coordinate sstem, i.e. at x =. The problem is two dimensional and the field that results after transients have decaed is simpl ux = Gx,x q dx C and we approximate the Green s function as the free-space one since the question sas we can ignore boundar effects. So we have ux = q ln x x dx. Introduce the two polar coordinate sstems C x = r cosθ, = r sin θ, 3.6 x = r cosθ, = r sin θ, 3.6 and thus, upon simplifing the argument we find ur, θ = q = q = q = qa2 2 ln r + q lnr 2 + r 2 2rr cosθ θ r dr dθ ln r 2 + r2 r 2r 2 r cosθ θ r dr dθ [ ] ln r 2 + ln + r2 r 2r 2 r cosθ θ r dr dθ 3.62 ln + r2 r 2r 2 r cosθ θ r dr dθ 3.63 Let β = r /r and ψ = θ θ so that dr = rdβ and dψ = dθ ur, θ = qa2 2 ln r + qr2 /r β ln[β 2 + 2β cosψ] dβdψ 3.64

2 William J. Parnell: MT3432. Section 3: Green s functions in 2 and 3D 7 where we note that the limits of the θ integral simplified because we could write θ θ = θ + θ and since the integrand is periodic with period, the first and last terms on the right hand side cancel. We now use the result given in the question, so that if r > a, the β integral runs from to a/r < and therefore is zero. Therefore for r > a, ur, θ = qa 2 /2 lnr. If r < a, the β integral runs from to a/r > so we write it as qr 2 + /r β ln[β 2 + 2β cosψ] dψdβ. The first term integral between and is zero using the result 3.59, the second, again using the result 3.59 again qr 2 /r [ β ln β dβ = qr 2 β2 4 + ] a/r 2 β2 ln β = qa2 4 + qa2 qr2 [ln a ln r] where we have evaluated this using integration b parts. Finall when combined with the first term in 3.64, this reduces to and therefore we have q 4 r2 a 2 + qa2 2 ln a q ur, θ = 4 r2 a 2 + qa2 ln a, r a, 2 qa ln r, r a. which we note is continuous at r = a on the boundar of the forcing disc C. Finall we just have to check that the integral at infinit from Green s identit does not diverge as described above. If it diverges then the solution 3.4 is not the correct one and our assumptions from the outset are wrong. So we must see if 3.52 holds. Well it clearl does since with u = qa 2 /2 lnr, lim R u r ln r u r r=r qa 2 = lim lnr r ln r R 2 r r=r 3.67 = so the boundar term is not just zero in the limit, it is identicall zero. We plot the solution in figure 8

3 William J. Parnell: MT3432. Section 3: Green s functions in 2 and 3D 7 Figure 8: Plot of the stead state temperature distribution ux, in a domain where boundaries ma be neglected with the heat source inside the circular domain x <. Here we have a = q =. Example 3.5 Consider the stead state heat distribution governed b 2 u = on the semi-infinite domain D = { < x <, } subject to the inhomogeneous boundar condition ux, = hx. Use the semi-infinite Green s function to determine the solution to the problem in integral form and determine the solution explicitl when { qx, x a, hx = 3.69, otherwise for a, q R, with a >. Since we have a semi-infinite domain let us emplo the semi-infinite Green s function that we determined in 3.57 above. We can use the expression 3.39 derived in section with Qx =, i.e. ux = hx n x Gx,x ds. 3.7 D

4 William J. Parnell: MT3432. Section 3: Green s functions in 2 and 3D 72 Firstl, we note that the dumm variable of integration is x. This is important! The boundar should be partitioned as D = D D, i.e. one along the boundar =, D and one from infinit, D. See figure 9. We assume that the boundar integral along D, i.e. the boundar contribution from infinit is zero. But we will check this later! The orientation of the integral is anti-clockwise with the normal pointing outwards. Therefore, since also hx = outside x a, and it is qx inside that interval, we have ux = qx n x Gx,x dx. 3.7 = where we note that s = x on D and this integral is along = figure 9. Next, we note that n =, so that x Gx,x = Gx,x Next we use the Green s function that we determined in Example 3.3, in the form Gx,x = ln x x ln x x where x = x, with >, and x = x,. We have Gx,x = Gx,x. Therefore Gx,x = x x x x and we have to evaluate this on =. Gx,x = = x x x x = x x Finall then ux = q x x x ds where we have taken q outside the integral. You will hopefull remember how to evaluate integrals of this form from the first ear! What we have to do is to re-write it b using the standard trick of adding zero, i.e. ux = q = q = q x x x dx x x x x x dx x x x x dx + qx x x dx. 3.75

5 William J. Parnell: MT3432. Section 3: Green s functions in 2 and 3D 73 D R n D x Figure 9: Figure illustrating the decomposition of the boundar D into a contribution from D, i.e. from infinit as R and one on =, i.e. D. The first integral is q and let x x x x dx = q [ lnx x ] a The second integral is qx = q = q ln x x dx = qx x + a 2 + 2, 3.76 [ lnx a lnx + a ], 3.77 x a x x 2 / 2 dx p = x x so that dx = dp and the lower and upper limits become x/ and a x/ respectivel: qx a x/ x x dx = qx dp 3.79 x/ + p 2 = qx [ ] a x/ p 3.8 x/ = qx [ ] a x x 3.8 = qx [ ] x + a x a 3.82 where the last step used the fact that is an odd function. So combing 3.78 and 3.82, the solution is ux = q x a 2 ln qx [ ] x + a x a x + a

6 William J. Parnell: MT3432. Section 3: Green s functions in 2 and 3D 74 Figure : Plot of the stead state temperature distribution ux, in the semi-infinite domain with boundar distribution hx given b Let us check that this recovers the right behaviour as. The first term tends to zero so we eliminate that straight awa from our investigations. The behaviour of the second term is a little more subtle. Let us fix q = a = as in Figure. We have to consider three different domains for x. We stud Fx, = x [ ] x + x in the limit as + i.e. from above. We should find that lim Fx, = hx First consider x <. In this case the arguments of become x + lim, x lim, so both s ield /2 and the respective contribution cancels, as is required for hx.

7 William J. Parnell: MT3432. Section 3: Green s functions in 2 and 3D 75 Then Second consider < x <. In this case the arguments of become lim lim x + lim, x lim, x = /2, x = / and so their respective contributions add to get, for < x <, lim Fx, = x = x as is required for hx. Finall consider x >. In this case the arguments of become x + lim, x lim, so both s ield /2 and the respective contribution cancels, as is required for hx. So we have indeed proved that as required. lim Fx, = + { x, < x <,, otherwise 3.94 = hx 3.95 Finall we should show that the contribution from the integral at infinit is zero. This is rather difficult! But just involves taking limits of the solution and its derivative, in order to show that as r ur, θ r 2, and therefore the contribution from this integral is zero. u r r, θ r Some other slightl more simple examples involving boundar forcing are considered on Sheet 7. We did this example here so that ou could see all of the separate elements that can occur.