 Welcome to this lecture on Dirichlet boundary value problem on a disk in R2 for Laplace equation. The outline is as follows. We solve the Dirichlet boundary value problem on a disk using separation of variables method and we rederive Poisson's formula in this situation. For most part of this lecture it is going to be computation. So please stop at every slide and do the computations on your own. So Dirichlet boundary value problem on a disk in R2, let D of 0, R denote the disk having center at the origin and radius R. That is D of 0, R is set of all x, y in R2 such that x square plus y square is less than R square. Boundary of D of 0, R is a circle and is denoted by S of 0, R. For a given function f which is continuous on S of 0, R that is on the circle, we would like to solve the following Dirichlet boundary value problem. What we end up solving using this method of separation of variables, we will be able to justify only if this f is little more nice. What we are going to see is that we need some more conditions on f and which we can justify the separation of variables method that we are going to discuss. First of all does separation of variables method work? In separation of variables method we look for solutions to any equation in this context it is Laplace in equal to 0, having this separated form u of x, y is a function of x into a function of y. We can substitute in Laplace in equal to 0 and get to od is 1 each for x and y. But the boundary condition is u of x, y equal to f of x, y given on the circle. From here we cannot get any boundary conditions for x or y. Note that the domain of a function which is in this product form x of x into y of y is domain of x cross domain of y. Because x of x into y of y makes sense for every to pull x, y such that x belongs to domain of x and y belongs to domain of y. Since the domains of x and y will be intervals because x and y are functions of one variable the domain of the product will be a right angle in x, y coordinate system. Thus separation of variables method will not work for disc. Thankfully disc is a rectangle it looks surprising disc is not rectangle. But disc is a rectangle in some other coordinate system namely the polar coordinate system. Now the equation Laplace in equal to 0 needs to be transformed into r theta coordinate system. So Laplacian in polar coordinates the Dirichlet problem for Laplace equation in polar coordinates is given by these two equations. This is precisely the Laplacian in the polar coordinates. So we have done NF exercises on how the PDE transforms under change of coordinates. So I leave this as an exercise for you to check that Laplacian in polar coordinates is given by this operator which is here vrr plus 1 by r vr plus 1 by r square v theta theta. We are given u of x, y on the circle that will define a function capital F for theta in 0 to 2 pairs follows. F theta is nothing but this is a given F of r cos theta r sin theta. In polar coordinates x equal to r cos theta y equal to r sin theta. So this is how you get for points on the circle of radius r with center at origin. So what are the main steps in separation of variables method? Step one, two families of ODEs obtained from Laplace equation or any equation for which you are trying to do separation of variables method. Separation of variables method is not new to us. We have done many exercises based on this for wave equation. So look for solutions of the Laplace equation in polar coordinates in the separated form v of r theta as h of r into G of theta. The Laplace equation will give rise to two families of ODEs indexed by a single parameter lambda 1 for h and 1 for g. Equation for g will be supplemented with periodic boundary conditions which is G of 0 equal to G of 2 pi and G prime of 0 equal to G prime of 2 pi. The question is why? Why because what is our aim? We want to solve Laplace equation in the disk. That means the function should be continuous, differentiable, C2 etc. But that we are trying to solve in terms of h of r and G of theta. So there should be some requirement on G that is precisely this. Please explore this further and then find out the precise reason which I have only indicated why this condition is needed. At this stage we cannot infer any conditions for h or g from the Dirichlet boundary data. The boundary data will be used at the very end of the procedure. If you recall for the wave equation also one equation we have solved a boundary value problem. For the second equation we do not have boundary value problem we have if you want a partial initial value problem and still there were some constant to be determined that we used the other initial data to fix those constants. Same thing here. So step 2 obtaining non-zero solutions to the two families of ODE's. The BVP for G of theta turns out to be an eigenvalue problem. It turns out that only a countable number of such BVPs index by n belongs to n union 0 will have non-zero solutions. For each of the eigenvalues lambda n we need to solve the ODE for h of r. At the end of step 2 we have a countable number of non-zero functions h n of r into g n of theta. Step 3 is the proposal of a formal solution. As a linear superposition of h n of r into g n of theta n belongs to n union 0. A superposition of the functions h n r g n theta n belongs to n union 0 is proposed as a formal solution to the Dirichlet boundary value problem. The Dirichlet boundary condition will be used to determine the weights in the superposition. Finally, one needs to check that the formal solution is indeed a solution to complement all the hard work that we have done so far. So step 1 method of separation of variables looks for solutions of the form h of r into g of theta where g of 0 equal to g of 2 pi and g prime of 0 equal to g prime of 2 pi. The periodic property for g and g dash substituting in the Laplace equation in polar coordinates and then dividing both sides with h of r into g of theta gives us this equation. So rearranging the terms in this equation gives us this equation. I have taken the g terms to the right side and multiplied with r square. So I have got this. Now observe that LHS is a function of r only, RHS is a function of theta only. Therefore, such an equation can hold if and only if both functions are equal to a constant function. So it means that there exists lambda in r such that you have this equal to this and both of them equal to lambda. One of the tasks is to find all possible lambdas coming from separated solutions. This gives rise to 2 ODE's which are given by g double dash plus lambda g equal to 0 r square h double dash plus r h dash equal to lambda h. Thus we have the boundary value problem for g of theta given by g double dash plus lambda g equal to 0 and this periodic boundary conditions. And the ODE for h of r is given by r square h double dash plus r h dash equal to lambda h Dirichlet boundary conditions to be used later. Now let us try to solve this boundary value problem for g. So there is a parameter lambda. So the lambdas for which the BVP admits a non-zero solution are called eigenvalues and the corresponding non-zero solutions are called eigenfunctions. Let us start our search for eigenvalues and eigenfunctions. Note that the lambda can be 0 positive or negative. These are the only 3 possibilities for a real number. Let us solve the BVP with lambda equal to 0 and see whether it has a non-trivial solutions or not. So the BVP itself becomes this because the equation changes lambda equal to 0. So the g term is dropped g double dash equal to 0. Now general solution of the ODE g double dash equal to 0 is a theta plus b. Now apply these boundary conditions g of 0 equal to g of 2 pi if you put what we get is a equal to 0. Thus lambda equal to 0 is an eigenvalue and eigenfunctions are constants. The second boundary condition does not give us any information because g prime is actually a, a equal to a. So it is always true. So essentially the first boundary condition gave us a equal to 0 therefore eigenfunctions are g theta equal to b, b is any constant. Now let us look at boundary value problem for lambda positive. So once lambda is positive we can write it as mu square for mu positive for definiteness and also because we are dealing with the second order equation it is convenient to have somebody square here. So that is why mu square. Now general solution of the ODE is combination of cosine and sine functions a cos mu theta plus b sin mu theta. Now we have to determine the constants a and b such that these boundary conditions are satisfied and then ask if it is possible to choose the constants a and b at least one of the nonzero in which case we get eigenvalues and eigenfunctions. Therefore this function here satisfies these boundary conditions if and only if these two conditions are met I have just substituted into these expressions this formula. So we get these two equations coming from these two conditions. Now it is always a good idea to write it as a linear system. So the two equations we have on the last slide let us write it as a linear system and we are looking for nonzero solutions of this system because that will give us eigenvalues eigenvectors. It has non-trivial solutions if and only if this determinant is 0 if the determinant is nonzero that implies a, b is 0, 0. So that is not a situation we want. So therefore we want this determinant to be 0 which will give us cos 2 mu pi equal to 1 that means mu is a natural number. So therefore we have a sequence of eigenvalues. What are the eigenvalues? They are mu square. So mu is n here so any natural numbers so eigenvalues are squares of natural numbers and what are the eigenfunctions? They are generated by cos n theta and sin n theta. For each n it is a linear combination of cos n theta and sin n theta that will be an eigenfunction. Now let us enquire into negative lambdas whether those BVPs have a non-trivial solutions or not. So we can write lambda equal to minus mu square where mu is positive and the equation is here. These are the boundary conditions. So general solution of the above ODE is a combination of exponentials which is here and let us substitute these boundary conditions. We get these relations to understand whether this has a non-trivial solution or not. It is always a good idea to write it as a linear system which we do here. Now we enquire into whether the determinant is 0 or nonzero. If it is 0 only trivial solutions that means no negative eigenvalues. If determinant is 0 then we have non-trivial solutions then there are negative eigenvalues. So above system has non-trivial solutions if and only if the determinant is 0 which comes out to be this e power 2 mu pi plus e power minus 2 mu pi equal to 2 and it looks like a number plus 1 by that number equal to 2 for which alpha equal to 1 is the only solution. Since mu is positive it follows that this cannot be equal to 2 because alpha equal to 1 means mu is 0 but mu is not 0 therefore this equation is not satisfied. It means that there are no negative eigenvalues. So let us summarize the eigenvalues and eigenfunctions to the BVP for G of theta. They are indexed by N union 0 natural numbers union 0. Lambda naught equal to 0 is an eigenvalue with an eigenspace consisting of constant functions. Lambda n equal to n square with eigenspace spanned by cos n theta and sin n theta. Now it is time to solve the equation for H of r which is here for each lambda equal to lambda n N in natural numbers union 0. Lambda naught equal to 0 general solutions to the above E is a log r plus b where a and b are real numbers. Recall that our goal is to solve the boundary value problem on the disk in particular the function the solution that we are planning to get should be bounded at the origin it should be bounded on the disk but bounded at the origin in particular. Therefore log r term is not suitable for that therefore a must be 0. Thus we obtain the following solution of Laplace equation H naught of r into G naught of theta. Remember G naught of theta was also constant when lambda naught equal to 0 H naught of r is also constant because a is 0 there is only b here. So we can take that constant to be 1 anyway we are going to take a linear superposition of these numbers. So you keep it 1 or 2 is the same so keep it 1. For each n in N union 0 the ODE for H with lambda n equal to n square is given by this equation. The ODE is of Cauchy Euler type which reduces to a constant coefficient ODE by a change of variable s equal to log r and solving is left an exercise or nowadays people also know what kind of solutions are there for this equation they try to wait look for solution of the form r power k and then you can solve like that also. So the general solution of the above ODE is given by a r power n plus b r power minus n where a and b are real numbers. Once again recall that our goal is to solve the BVP on the disk containing origin therefore we are looking for solutions which are bounded and hence this b should be 0. Thus we obtain the following solution h n r into g n of theta given by r power n from h n and this is from g n. Now let us propose a formal solution to the Dirichlet boundary value problem using superposition principle by V of r theta equal to a naught by 2 plus summation n equal to 1 to infinity r power n a n cos n theta plus b n sin n theta. If you recall this is a combination of the function 1 which is coming from h naught of r into g naught of theta and this sum n here is precisely h n of r into g n of theta. So we are taking a combination and the constants in the combination are taken a n and b n. So the unknown coefficients a 0 a n and b n has to be determined using the boundary condition because that is the only condition we have not used yet. That means V of r comma theta that is r equal to capital R should be f of theta and that should be equal to this expression with r equal to capital R. This is how we would like to find the constants a n b n a naught such that the function f of theta is given by this Fourier series. So choose the constant r power n a n into r power n b n such that they are the Fourier coefficients of f of theta and of course we have this expression for a n and b n. So this finishes the construction of a formal solution to the Dirichlet BVP. So formal solution to the Dirichlet BVP is given by this I have substituted the coefficients a n b n into the series that we have proposed. So now this has no a n b n everything in terms of f and this has a cos and sin terms inside. So is it a solution to Dirichlet BVP that is a question now. To show that the formal solution given here is a solution we need to prove that the series here which this is just one term there is a series here that series converges in other word this makes sense and defines a function if such a thing happens call it v of r theta and then we have to show that v of r theta is a C2 function we need to show that infinite series can be differentiated two times and the resultant series converge then we have to show that the boundary condition is satisfied v of r theta equal to f theta holds these three things we have to do. So from now onwards you can just see the video once and you can forget the details because some of the details I have not mentioned very clearly for which I have given you a reference but we should be happy that we have got a formal series here. Regulus analysis of course I have given from now onwards. So questions of convergence of the infinite series appearing in this formal solution namely this one are better handled using its original avatar which is this where a and a has this formula b and has this formula. So on formally differentiating the formal solution we get v r to be like this v r to be like this v theta theta should be like this. So every time you differentiate you pick up some n so rn n is coming here here n into n minus 1 that is n square term. So in second order derivatives you have n square terms and in first order derivatives you have n terms when you have no derivative there is no n as a coefficient here of course r power n cos n theta they are all there. So substituting these expressions for v r v r r v theta theta which are given here formally you add them up with the corresponding weights which weights the one which are in the Laplace operator this one. So v r r plus multiply v r with 1 by r v theta with 1 by r square you see that you get 0. So formally we have shown that when the series on the previous slide makes sense they add up and in this way to get that v of r theta is a solution to Laplace equation. So it reminds to prove that all the series are convergent and of course the boundary condition is satisfied that we have to prove. So once the series are convergent we have just shown that Laplace and v equal to 0 is satisfied. So convergence of the series for v r r guarantees the convergence of the rest of the things because of the estimate that we are going to prove for this. Reason I have explained that there is a n square here there is no n here there is n here there is n square here. So this and this we have similarly whereas these behave slightly better in fact more better than the second order derivatives. Look at this estimate this is the general term in the series for v r r that is less than or equal to this n into n minus 1 that is n square minus n that is less than or equal to n square r power n minus 2 as it is and cos n theta sin n theta modulus is less than or equal to 1 so you have mod n plus mod b n and that is less than or equal to this quantity in view of this relations. So please do this computation by yourself. This series is convergent by let us say ratio test for all r less than capital R. So it follows that the series in the definition of v of r theta along with the series obtained after differentiating the one for v are all uniformly convergent for all r less than or equal to r naught where r naught is strictly less than r and for every such r naught. Therefore term by term differentiation of a series for v of r theta is valid. Our proof can be continued to show that v is in fact an infinitely differentiable function in the open disk d of 0 comma r. A reference is booked by Weinberger on PDE with complex variables and transform methods. I have taken this material from this book all these details you can see in more details in this book. So we will use the maximum principle to show that the boundary condition is satisfied. So you can think this is an application of maximum principle. So we need to assume conditions on F which guarantee that this Fourier series on the right hand side converges to F uniformly that is needed. A sample condition is this, F is continuous periodic, F dash is square integrable on 0 to pi. Of course there are many more such a sufficient conditions. Let S k denote the kth partial sum in the series for F that is this you truncate this after k times. So for k bigger than L S k theta minus S l theta is given by this. Let vk denote the kth partial sum in the series of v for k bigger than L vk minus vl has this expression. It is a harmonic function because each of the inside terms is harmonic function is a finite sum therefore it is a harmonic function. And it equals S k theta minus S l theta on the boundary when r equal to r and vk minus vl is a c2 function on the disk and it is continuous off to the boundary of the disk. In fact vk minus vl is c infinity on r2. So by weak maximum principle we have this maximum in the domain or omega closure is less than or equal to maximum on the boundary. Since S k is a Cauchy sequence in c of 0 to pi so and it converges to capital F right that is why it is a Cauchy sequence and hence vk will be a Cauchy sequence in this domain 0 less than or equal to r less than or equal to r theta belongs to 0 to pi which is the closed disk. So therefore vk converges uniformly in the closed disk and hence the limit has to be a continuous function which is continuous off to the boundary by the property that it is continuous on d of 0 for v is continuous and v of r theta equal to f theta is satisfied. Now let us re derive Poisson's formula. The summation and integral can be interchanged in this expression since the series is uniformly convergent for r less than or equal to capital r. So this is equal to this in terms of the complex exponentials and on simplification it is the term in the brackets is like this. Now the term within the square brackets here is looking like 1 plus z by 1 minus z plus z bar by 1 minus z bar which can be computed to be this. Please do this computation. So now I am going to substitute what is my z that is r by r e power i into tau minus theta and I get this expression. So solution to Dirichlet boundary value problem in polar coordinates is given by this for r strictly less than r. r equal to r this is not meaningful you should not do that r less than r. In x y coordinates the above formula transforms to the well known Poisson's formula. It is valid on the disk D of 0 comma r. Let us summarize what we did in this lecture. Dirichlet boundary value problem on a disk is solved using separation of variables method. Knowing that disk is a rectangle in polar coordinates the BVP was transformed to polar coordinates. The problem becomes suitable for using separation of variables method since the series expansion could be summed up Poisson's formula was obtained as a byproduct. Thank you.