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Title: Laplace’s Equation
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Series: Partial Differential Equations
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Chapter: Laplace’s Equation
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YouTube-Title: Partial Differential Equations 2 | Laplace’s Equation
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Subtitle in English
1 00:00:00.480 –> 00:00:05.680 Hello and welcome back to partial differential equations, the video series where we look at
2 00:00:05.680 –> 00:00:11.840 some specific PDEs and try to develop a theory for them. And in today’s part two,
3 00:00:11.840 –> 00:00:18.400 we start talking about a very important PDE called Laplace’s equation. This one occurs a
4 00:00:18.400 –> 00:00:24.000 lot in applications, especially in physics when you have something like a potential.
5 00:00:24.000 –> 00:00:28.160 However, as always, before we go into the definition, I first want to thank all the
6 00:00:28.160 –> 00:00:33.760 nice people who support the channel on Steady, on YouTube or via other means. And as a reminder, you
7 00:00:33.760 –> 00:00:39.520 can use the link in the description to download additional material for the videos. For example,
8 00:00:39.520 –> 00:00:45.280 there you can find the PDF versions and quizzes for the videos. And then I would say let’s start
9 00:00:45.280 –> 00:00:52.160 with the topic of today by defining the so-called Laplacian. As already mentioned in the last video,
10 00:00:52.160 –> 00:00:59.520 this one is denoted by capital delta and also often called Laplace operator. The term “operator” fits
11 00:00:59.520 –> 00:01:05.040 because it acts on functions and a new function comes out. And here we have a function u defined
12 00:01:05.040 –> 00:01:12.880 on Rn or on an open subset omega in Rn. And then the output space is just the real number line as
13 00:01:12.880 –> 00:01:20.080 always. And now the Laplace operator acted on u has a nice definition because it just uses the second
14 00:01:20.080 –> 00:01:26.960 order partial derivatives of u. More precisely, it uses the one with respect to just one variable,
15 00:01:26.960 –> 00:01:34.080 for example, x1. And then the other ones are just added until we reach xn. So in short,
16 00:01:34.080 –> 00:01:40.640 we just have a sum with n terms where n is the dimension of the input space of u. And in the case
17 00:01:40.640 –> 00:01:46.320 you want to write the partial derivatives with the multi-indices, you would say you have D alpha of
18 00:01:46.320 –> 00:01:53.120 u where alpha is always of order two with an additional condition. Namely only one
19 00:01:53.120 –> 00:01:59.680 entry alpha j is non zero. So it has the same meaning as before. We use the second order
20 00:01:59.680 –> 00:02:06.800 partial derivatives but not the mixed ones. And now by using this operator we can define Laplace’s
21 00:02:06.800 –> 00:02:13.200 equation as a PDE. And you might already know that it just says that the Laplacian of u should
22 00:02:13.200 –> 00:02:19.920 be equal to zero. So we search for the functions u that satisfy this equation on the domain
23 00:02:19.920 –> 00:02:24.960 omega. And here I can immediately tell you an application in physics. For example, you want
24 00:02:24.960 –> 00:02:30.800 to calculate the electric potential when there is no charge around. So at every point where there
25 00:02:30.800 –> 00:02:37.200 is no source, no charge, the electro potential should satisfy Laplace’s equation. And in fact,
26 00:02:37.200 –> 00:02:43.120 at the end of this video, we can immediately write down some solutions of this PDE. However,
27 00:02:43.120 –> 00:02:49.200 first I should tell you that the solutions of this equation have a special name in mathematics. More
28 00:02:49.200 –> 00:02:55.280 precisely, we want to have solutions on an open set omega and where the partial derivatives of
29 00:02:55.280 –> 00:03:03.600 second order are still continuous functions. So in short, this means we search for function u in C2.
30 00:03:03.600 –> 00:03:07.920 So you know this the common notation for continuously differentiable functions up
31 00:03:07.920 –> 00:03:15.200 to order two. So obviously for these functions we don’t have a problem applying the Laplacian and in
32 00:03:15.200 –> 00:03:21.920 the case we get out zero we speak of a harmonic function and you see it’s not hard at all to find
33 00:03:21.920 –> 00:03:28.800 examples for harmonic functions because all the constant functions satisfy Laplace’s equation. These are
34 00:03:28.800 –> 00:03:34.560 not so interesting. So let’s search for some more harmonic functions and maybe the ones with some
35 00:03:34.560 –> 00:03:42.320 symmetry. I say that because a radial symmetry allows us to completely solve Laplace’s equation.
36 00:03:42.320 –> 00:03:48.080 So for the harmonic function u, we can take out the origin from the domain. So we consider it
37 00:03:48.080 –> 00:03:55.520 from C2 of R N without zero. So in the case n is equal to two we can simply sketch the whole
38 00:03:55.520 –> 00:04:02.720 thing which means we have the whole space except for one point. So seeing that in the application
39 00:04:02.720 –> 00:04:07.600 you could say that we have a point charge at the origin and now we want to see what happens
40 00:04:07.600 –> 00:04:14.080 outside and there it’s quite natural to assume a radial symmetry which means that the value of
41 00:04:14.080 –> 00:04:19.840 u only depends on a distance to the origin. Hence in our two-dimensional picture here,
42 00:04:19.840 –> 00:04:26.400 it means that for a given circle, we find the same value for each point on it. And in the formula,
43 00:04:26.400 –> 00:04:32.800 this would mean that u of x stays the same if you change the direction of x, but not the length of
44 00:04:32.800 –> 00:04:40.000 x. Which also means that we already get all the values just by looking at one direction. So we can
45 00:04:40.000 –> 00:04:47.600 put these values into a real function phi that now only depends on the norm of x. And clearly what we
46 00:04:47.600 –> 00:04:55.440 mean here is the standard Euclidean norm in Rn. And obviously this function phi only has to be defined
47 00:04:55.440 –> 00:05:03.760 on the real interval from 0 to infinity and the values are the same as for u. So we land in r
48 00:05:03.760 –> 00:05:09.840 and moreover a common thing would be to denote the radial variable by r. Hence now everything
49 00:05:09.840 –> 00:05:16.720 is just about the function phi of r. So you see if we already assume that u is harmonic we need
50 00:05:16.720 –> 00:05:23.040 to apply the Laplace operator to this composition of two functions and then we definitely know the
51 00:05:23.040 –> 00:05:30.000 conditions that phi has to satisfy in this case. So again we assume that u is harmonic.
52 00:05:30.000 –> 00:05:36.640 So summing up these partial derivatives gives us zero. Therefore let’s first start with the
53 00:05:36.640 –> 00:05:44.800 first order partial derivative del u del xj. There we just have a chain rule because we have this
54 00:05:44.800 –> 00:05:52.640 composition of two functions. More precisely we have phi of the square root of the squares added.
55 00:05:52.640 –> 00:05:59.040 So the Euclidean norm in Rn is not so complicated and we can definitely form the derivative. So the
56 00:05:59.040 –> 00:06:06.880 first thing we get here is phi prime of x in the norm and then times the derivative of the square
57 00:06:06.880 –> 00:06:14.880 root times the inner derivative 2xj. So we have 1 / 2 * the square root which we could write as the
58 00:06:14.880 –> 00:06:22.400 Euclidean norm again time 2xj. So the good thing is we can already cancel the two and the whole thing
59 00:06:22.400 –> 00:06:29.280 is quite compact which helps us because we have to calculate a derivative again also not so hard
60 00:06:29.280 –> 00:06:35.360 because we just have to calculate with the product rule. So we immediately get here the second
61 00:06:35.360 –> 00:06:42.640 derivative of phi and please don’t forget we also get out the inner derivative again and multiplying
62 00:06:42.640 –> 00:06:48.880 that with the factor we already have gives us the same thing squared and then finally the product
63 00:06:48.880 –> 00:06:55.840 rule tells us that we have a plus now and there if you want you can actually do a quotient rule here
64 00:06:55.840 –> 00:07:04.240 for the remaining term which means we just have the denominator squared. So the norm of x^2 and in
65 00:07:04.240 –> 00:07:11.040 the numerator we multiply this with the derivative of the former numerator which is just 1 and then
66 00:07:11.040 –> 00:07:18.080 we have minus xj times the same inner derivative as before. Hence you immediately see here it’s
67 00:07:18.080 –> 00:07:24.880 helpful to multiply with the norm of x in the numerator and denominator because then we find
68 00:07:24.880 –> 00:07:30.960 the power three on the bottom power two here and there. So actually it’s quite nice and now we can
69 00:07:30.960 –> 00:07:37.760 form the whole sum and we already know it’s equal to zero and there you might already recognize the
70 00:07:37.760 –> 00:07:44.560 sum here for the first term gives us the norm of x squared in the numerator as well. So this nicely
71 00:07:44.560 –> 00:07:51.760 cancels exactly what we want to have. And then for the second part here we also have the norm of
72 00:07:51.760 –> 00:07:58.880 x^2 there. However, please note that we have a sum over n terms, which means that this constant term
73 00:07:58.880 –> 00:08:06.800 here actually occurs n times. And there we have it. This is what Laplace’s equation tells us for the
74 00:08:06.800 –> 00:08:16.960 function phi. The second derivative plus the first derivative * n -1 / the norm of x is equal to
75 00:08:16.960 –> 00:08:24.240 zero. In other words, what we get out here is an ordinary differential equation for phi. So let’s
76 00:08:24.240 –> 00:08:31.680 summarize that as our important result here. A radial symmetric harmonic function u on a pointed
77 00:08:31.680 –> 00:08:40.000 domain can be written with a function phi that satisfies this ODE. And we can write this ODE as
78 00:08:40.000 –> 00:08:47.440 the second derivative of phi at the point r plus the first derivative of r which has the factor n
79 00:08:47.440 –> 00:08:56.960 -1 / r and this should be equal to zero for every r between 0 and infinity. So this is quite nice
80 00:08:56.960 –> 00:09:02.960 and with our knowledge of ordinary differential equations we can immediately solve this one. In
81 00:09:02.960 –> 00:09:09.440 fact it’s a linear one which we can solve with an integrating factor. So maybe it’s easier to see
82 00:09:09.440 –> 00:09:16.880 if you write it as f of r as the first derivative of phi because then you just see that we actually
83 00:09:16.880 –> 00:09:23.920 just have a first order od to solve. And you might remember that the integrating factor just needs
84 00:09:23.920 –> 00:09:30.480 the anti-derivative of the factor in front of f of r which is quite simple here because it’s just
85 00:09:30.480 –> 00:09:36.800 the logarithm function. And since I don’t want to confuse, I will use this common ln notation
86 00:09:36.800 –> 00:09:43.280 for the natural logarithm. And now the advantage here is that we can use the logarithm rules, which
87 00:09:43.280 –> 00:09:49.600 means we can pull in this factor as an exponent. And this is quite helpful because the integrating
88 00:09:49.600 –> 00:09:57.440 factor we need to solve the ODE is always given as e to the power of the anti-derivative. So in this
89 00:09:57.440 –> 00:10:04.800 case, it’s simply r to the power n minus one. So this is the factor we multiply on both sides and
90 00:10:04.800 –> 00:10:10.560 then we can just use the standard product rule. Hence the whole differential equation can just
91 00:10:10.560 –> 00:10:21.200 be written with one derivative with respect to r and we use that on the function r ^ n -1 * f of r.
92 00:10:21.200 –> 00:10:27.200 And now in this form the order E is quite easy to solve because we see that the function inside has
93 00:10:27.200 –> 00:10:34.160 to be equal to a constant on the interval from 0 to infinity which then implies that f of r is
94 00:10:34.160 –> 00:10:42.320 equal to this constant c1 divided by the function r ^ n minus one. And obviously this inverse can
95 00:10:42.320 –> 00:10:50.480 also be written as r ^ 1 - n. And with that we are almost done because now we only have to integrate
96 00:10:50.480 –> 00:10:57.760 this function f of r to get our original function phi. And this already means that we also get in a
97 00:10:57.760 –> 00:11:05.120 second constant c2. Moreover for this integration we also have to distinguish two different cases
98 00:11:05.120 –> 00:11:11.680 here. You can see that in the formula of our function because the case n is equal to 2 gives
99 00:11:11.680 –> 00:11:18.720 us the function 1 / r. And there we already know integrating that gives us the natural logarithm
100 00:11:18.720 –> 00:11:27.440 function. Hence in this special case we have c2 plus c1 * ln of r. And this is the only special
101 00:11:27.440 –> 00:11:34.480 case because for a larger n we just have a general formula for the anti-derivative. Namely as you
102 00:11:34.480 –> 00:11:43.200 know we just have to increase the exponent exactly by one. So we have r ^ 2 - n with the same factor
103 00:11:43.200 –> 00:11:49.920 in front. But honestly, we could push this factor into the constant c1. And there we have it. These
104 00:11:49.920 –> 00:11:58.080 are all possible radial symmetric solutions of Laplace’s equation on the domain Rn without zero.
105 00:11:58.080 –> 00:12:03.600 And often in applications, as already mentioned, the three-dimensional case is important. And you
106 00:12:03.600 –> 00:12:09.840 see we just have the function 1 / r here. And soon we will see that this is actually
107 00:12:09.840 –> 00:12:16.480 what we call a fundamental solution of our PDE. Therefore, I would say let’s continue with that
108 00:12:16.480 –> 00:12:25.400 in the next video. So, I really hope I meet you there again and have a nice day.
109 00:12:25.400 –> 00:12:35.200 Bye-bye.
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Date of video: 2025-05-21
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Last update: 2025-09
