 Welcome to this lecture, starting from this lecture we are going to study qualitative analysis of wave equation. So far we have done the quantitative analysis for the wave equation namely we have solved Cauchy problems, initial boundary value problems associated to the wave equation. So in today's lecture we are going to discuss a special property of solutions to wave equation in one dimensions, it is known as parallelogram identity. The outline for today's lecture is first we show that solutions to homogeneous wave equation satisfy parallelogram identity, then we show that C2 function satisfying parallelogram identity is indeed a solution to the homogeneous wave equation and we apply parallelogram identity and solve a few problems. So solutions to homogeneous wave equation satisfy parallelogram identity definition of a characteristic parallelogram. A parallelogram in the xt plane is said to be a characteristic parallelogram if each of its sides lies along a characteristic line recall that there are two families of characteristic lines for wave equation. They are x minus ct equal to constant and x plus ct equal to constant these are the two families. So let parallelogram pqrs be a parallelogram in the xt plane with sides pq, qr, rs and sp then parallelogram pqrs is a characteristic parallelogram if each one of the sides pq, qr, rs, sp lies along some member of one of the two families of characteristic lines. The picture is here pqrs the side pq lies on this line x minus ct equal to constant qr lies on x plus ct equal to constant rs lies on x minus ct equal to constant and sp lies on x plus ct equal to constant. So this is a characteristic parallelogram because each of its sides lies on some characteristic line. So we have this theorem suppose pqrs is a characteristic parallelogram with the line segments pr, pr and qs as its diagonals. This is just to fix this kind of a picture pr and qs are diagonals. So in principle q can be here and s can be here but we are going to say that without loss of general let us assume pqrs are described in this anticlockwise manner the vertices of the parallelogram after all these are only a description naming. Let u be a function having this form u of xt is equal to f of x minus ct plus g of x plus ct for some functions fg defined on r. So no assumptions on f and g what all we need is f and g are just functions defined on r then this automatically defines a function on r2 u of xt for xt belongs to r2. Conclusion is the values of u at the vertices pqrs of the parallelogram that is a characteristic parallelogram they satisfy the parallelogram identity u of p plus u of r equal to u of q plus u of s. So this is how the characteristic parallelogram looks like. So u of p plus u of r is equal to u of q plus u of s. So without loss of generality assume that the side pq lies along we have to set up some notations. So pq lies along this characteristic line x minus ct equal to k1 and the vertices are as in this picture namely they described in this anticlockwise manner just to set up notations and therefore qr lies on some member of the characteristic lines family and of course it has to be from the other family x plus ct equal to lambda l2 there is some number l2 there is some number k2 such that rs is along this line x minus ct equal to k2 there is a number l1 such that sp lies along x plus ct equal to l1 that is what precisely we are assuming. So without loss of generality the characteristic parallelogram may be described as follows the side pq lies along the characteristic line x minus ct equal to k1 for some k1 the vertices are described in the anticlockwise manner pqrs is a characteristic parallelogram therefore there are characteristic lines along which the sides of pqrs lie in other words there are numbers l1 l2 k2 such that the sides sp, qr and rs lie along the characteristic lines which are described here we already saw this in the picture. Since u has this form that u of xt equal to f of x minus ct plus g of x plus ct we get u of p equal to f of k1 plus g of l1 because p lies on x minus ct equal to k1 and x plus ct equal to l1. Similarly u of q equal to fk1 plus gl2 u of r is f of k2 plus g of l2 and u of s equal to f of k2 plus g of l1 from the above set of equality is the parallelogram identity for us you can easily check that u of p plus u of r is equal to u of q plus u of s. So a remark on the theorem recall that the general solution to the homogeneous wave equation utt minus e square u of xx equal to 0 is given by u of xt equal to f of x minus ct plus g of x plus ct where f and g are c2 functions defined in r. Therefore any c2 solution of the homogeneous wave equation satisfies parallelogram identity for every characteristic parallelogram. Recall parallelogram identity is stated only for characteristic parallelograms. So the next result asserts the equivalence of being a solution to the wave equation homogeneous wave equation and satisfying the parallelogram identity for every characteristic parallelogram. In other words c2 function satisfying parallelogram identity is a solution to the homogeneous wave equation. So let u from r to to r be a twice continuously differentiable function for every characteristic parallelogram pqrs with the line segments pr and qs as its diagonals the parallelogram identity holds. Conclusion u solves the homogeneous wave equation which is utt minus c square u of xx equal to 0. So proof of the theorem we are given that parallelogram identity holds for every characteristic parallelogram. Main idea is to cleverly construct useful characteristic parallelograms. This is a standard idea in mathematics whenever you are given wealth of information like here something holds for every characteristic parallelogram. If you want to use it you have to really exploit it by cleverly making choices. So it is easy to verify that these points pqrs are vertices of a characteristic parallelogram. In fact we have derived these points how they should look like and then wrote down here. So it is easier if you look at the picture. So let us take the point p as xi tau and this line is given by x minus ct equal to constant since this point lies on it that constant has to be xi minus c tau. So like this. Now I am going to consider q which is of the type xi plus s comma something I can determine what that point is using this equation that is why I get q. Similarly I propose s is like xi minus r comma something and that something can be determined by using this equation x plus ct equals xi plus c tau. I get s once I know s I can write down the equation of this characteristic line passing through this point which is this equation. Similarly from q I can write down the characteristic line passing through q the other one one I already know. So this is from the other family and I see where they intersect I get this point that is how the vertices were determined. So why is it a clever choice we will see it on the next slides in the proof. So this is the picture that we have for pqrs and they are actually vertices of a characteristic parallelogram. So we are now in a shape to apply the parallelogram identity u of p plus u of r equal to u of q plus u of s. So it can also be written as or rewritten as this uq minus up equal to ur minus us. Now notice what is u of q minus u of p it is a value of u at this point what is this point it is actually xi comma tau plus s into 1 comma 1 by c and when s equal to 0 I am at p. So this is some kind of difference of u values of u along this direction 1 comma 1 by c. Similarly this u of r minus u of s you see s and s by c. So the point r is nothing but this point s plus s times 1 comma 1 by c. So that is also a variation or difference in this direction 1 comma 1 by c. So if you divide this differences with s divided by small s which is this s and then take the limit as s goes to 0 what we get is a directional derivative of u in this direction 1 comma 1 by c at the point p. Similarly if you look at ur minus u s divided by small s this s then we get the directional derivative of u at this point s in the direction 1 comma 1 by c. So just substituting for pq r s we get this. Now you look at this this is a difference quotient. Difference quotient when you are trying to compute the directional derivative of u in the direction 1 comma 1 by c at the point xi tau. Similarly this also when you are trying to compute the directional derivative of u at this point xi minus r comma tau plus r by c which is the point denoted by capital S in the parallel graph in the direction 1 comma 1 by c. So passing to the limit yields the directional derivatives why the limit exists because we are given that the function is C2 therefore all partial derivatives of order 1 exist. So we can compute using any formula that you like so we get this. So if you expand what is this? Gradu is u x comma ut u x comma ut is the gradient dot 1 comma 1 by c that will give you u x plus ut by c. So we get this at the point xi tau similarly the RHS. So derivative in the direction, directional derivative in the direction of 1 comma 1 by c is nothing but this particular combination of the partial derivatives. So rewriting what we have here we get this. Now if you look at this is the point p this is the point s this is also like a difference quotient once you divide with R but in which direction this suggests xi tau this is xi tau plus minus r comma r by c. R is positive therefore plus r comma minus 1 comma 1 by c. As far as the direction goes it is 1 comma minus 1 by c. So passing to this limit as R goes to 0 we get the directional derivative in the direction 1 comma minus 1 by c which is here the first one here. This is the directional derivative in the direction 1 comma 1 by c minus 1 by c of this quantity which is there here for which we have the difference quotients here when you divide with R. So once you expand you get dou 2 u by dou x square minus 1 by c square dou 2 u by dou t square at xi t equals 0 xi tau is an arbitrary point therefore u satisfies the wave equation at every point. Of course here we have used that when you expand you will get dou 2 u by dou x dou t and dou 2 u by dou t dou x they get cancelled because u is a c2 function mixed partial derivatives are equal. So the two theorems establish the following equivalence for a function which is c2 of r cross r in fact the same proof works with c2 of r cross 0 infinity the following statements are equivalent. u solves the wave equation is same as saying that on every characteristic parallelogram u satisfies the parallelogram identity. The second statement is meaningful even for a continuous function in fact you do not even need continuity but I am just putting because there is something nice to have of course we can even take discontinuous functions that is another thing. What I am saying is that this second part namely parallelogram identity makes sense this statement makes sense without any requirement of differentiability on u. This observation provides us a way to generalize the notion of solution whenever required we are going to discuss them later on. So let us look at some applications of parallelogram identity which is in solution of an IBVP with the Dirichlet boundary conditions. So using parallelogram identity solve the Darbu problem what is Darbu problem ut t minus u xx equal to 0 that is a wave equation posed in which domain t bigger than maximum of x comma minus x that is nothing but this t is greater than mod x maximum of x comma minus x is precisely mod x. In this domain we have to solve wave equation and we are given Cauchy conditions u is given to be so u is given to be phi here and u is given to be psi here and this is the domain in which we have to solve and we are given phi and psi to be c2 functions satisfying phi of 0 equal to psi of 0. So let us solve this problem. So what are the steps involved? First step is finding a suitable characteristic parallelogram suitable means useful. Second is use parallelogram identity and obtain a solution. Of course third thing still remains that we have to check that the solution that we obtained in 2 is indeed a classical solution. Okay let us look at the step 1 here first. Step 1 is to find a suitable characteristic parallelogram. These are the lines x equal to t and x equal to minus t. Both of them are characteristic lines. So let me pick up a point p here. I name it as xi tau and not x and t because I would like to use this notation of x and t in describing the lines. So this line is x minus t equal to xi minus tau and this line is x plus t equal to xi plus tau. So we call that as p let us call this as q r is the origin and this is the s and here we are given u equal to phi and here we are given u equal to psi. Therefore u at p can be obtained very easily. What we need to know is what is q what is s. So let us find out what is q and s. So pq the line pq the side pq lies on x minus t equal to xi minus tau. Therefore q coordinates are given by xi minus tau by 2 comma minus xi by 2. Ps lies on x plus t equal to xi plus tau but s also lies on x equal to t. So x component t component must be same. Therefore s is xi plus tau by 2 comma xi plus tau by 2. So we know the coordinates for q and s r of course is 0 0. Therefore u of yeah we have to find what is u of q. So u of q so u of q is here it is given in terms of psi. So psi of tau minus xi by 2 and u of s is given in terms of phi that is phi of xi plus tau by 2. And what is u of r? u of r is u of 0 0 and that is equal to phi of 0. Of course we have assumed that is equal to psi of 0 by assumption you may call compatibility condition. So step 2 is to get a solution apply parallelogram identity we already computed u of r u of q u of s. So therefore u of xi tau is equal to psi of tau minus xi by 2 plus phi of xi plus tau by 2 minus phi 0. So in terms of x t I simply replace xi tau with x t. So u x t is equal to psi of t minus x by 2 plus phi of x plus t by 2 minus phi 0. So we have obtained the solution. Now what remains is step 3 I have to check that the u given by this box the formula is actually a classical solution to the given problem. And that follows from our assumptions on phi and psi we assumed this in fact we only need the in fact we only need that what do we need phi and psi should be c 2 functions because I should be able to differentiate the expression for u 2 times and I need continuity only up to 0. Of course one can check this problem is also well posed. So let us look at problem 2. Here we are supposed to solve homogeneous wave equation with initial conditions and possibly a non-zero boundary condition general function h of t. We will use parallelogram identity to solve in some region of this domain. The domain in which we are interested in solving is this x positive t positive. From our prior experience we do know that in this domain which is determined by the line x equal to t namely x bigger than t the D'Alembert formula holds for the solution. So essentially we need to solve at a point which is above this line let us say a point here using parallelogram identity. So therefore let us look at how to solve that in this region where x is less than t this region. So this is the line x equal to t these are the axes. Let us take a point p let us denote by xi tau because we are going to use x and t to describe the equations of the characteristic lines. So what we do is just take this line which is parallel to this characteristic and this is the other one and this is another one. So q r s. So we know parallelogram identity gives us that u of p is equal to u of q plus u of s minus u of r. What is u of q? It is determined in terms of h because u is equal to h. What is u of s? u of s is given in terms of p because u is equal to p here and u of r also. So it looks like it does not depend on psi. So it means we are airing somewhere then when we look back p q r s is only a trapezium it is not a parallelogram forget about being a characteristic parallelogram. So this is the wrong picture tempting but wrong picture. So what is the correct picture? This is the line x equal to t started a point p which is xi comma tau. So this has to be a characteristic. This is the other characteristic. These are the characteristics. Now it looks okay q this point is r this point is s. We have to determine what these points are. We know u at q because u is prescribed here as h but we do not know what is u of r and u of s. That needs to be determined once again using the Dallambert formula because for which Dallambert formula holds. So if you call this as origin o let us call this r dash u of r is given in terms of 0 o and r dash similarly this also. So in which case we have to find out what are these points r dash and s dash to get solution at these points. So it is a two step process. So at this in this picture what we have is p q r s is a parallel characteristic parallelogram. Therefore u of p is equal to u of q plus u of s minus u of r by parallelogram identity. So now what we have to do is compute u of r u of s and substitute in this formula r dash s dash. So p q the side p q lies on x minus t equal to psi minus tau line. Therefore q is where x is 0 therefore 0 when x is 0 t is tau minus psi. Now let us look at p s that lies on x plus t equal to psi plus tau. Therefore the point s is psi plus tau by 2 psi plus tau by 2 because the point which lies on the line x equal to t. So x and t coordinates are same. Let us look at q r it lies on x plus t is equal to tau minus psi because it passes through the point q. Therefore the point r is given by tau minus psi by 2 comma tau minus psi by 2 because r is also a point which is lying on the line x equal to t. Therefore we can write down what are the values. Let us write 1 by 1 what is u of r? u of r is given by phi of 0 plus phi of tau minus psi by 2. I am using the Dalamboid formula for us psi is 0. So the coordinates of r dash are tau minus psi comma 0. Similarly u of s is equal to because s dash is psi plus tau comma 0. The value is phi of 0 plus phi of psi plus tau by 2. So we got r and s. So therefore u at the point psi comma tau is u of p. By parallelogram identity it is u of q plus u of s minus u of r. So that is nothing but h of tau minus psi plus phi of psi plus tau minus phi of tau minus psi by 2. So now let us switch to x t instead of psi tau because now we have a formula. So u of x t is nothing but h of t minus x plus phi of x plus t minus phi of t minus x by 2. Therefore let us write down the full solution. Full solution means we write down what is solution in x less than t? x greater than t in one place is u of x t equal to phi of x minus t plus phi of x plus t by 2 if x is greater than or equal to t h of t minus x plus phi of x plus t minus phi of t minus x by 2 if x t is bigger than x. So this is region 1, this is region 2. So this is for the region 1 and this is for the region 2. In region 1 it is given by D'Alembert formula. Now let us make some observations. First point is that u is smooth everywhere as smooth as the given function phi and h are in each of these regions 1 and 2 everywhere in the first quadrant except possibly on this line x equal to t. So let us examine what happens on the line x equal to t. So first part is continuity. Is it continuous at points of x equal to t? So from region 1 what we get is phi 0 plus phi of 2x by 2 and from region 2 what we get is h 0 plus phi of 2x minus phi of 0 by 2. So this is same as phi of 0 equal to h of 0. So this is one compatibility condition that we get. So continuity of this function demands that phi 0 must be equal to h 0. Similarly, let us check for c 1s. For that we need to write what is ux of x t of course u t of x t as well. So ux of x t is phi prime of x minus t plus phi prime of x plus t by 2 in the region x bigger than t minus h prime of t minus x plus phi prime of x plus t plus phi prime of t minus x by 2 in the region t bigger than x. So this is the region 1, this is the region. So therefore ux is continuous if and only if the limits from both the regions 1 and 2 as we approach x equal to t coincide. So what we have is phi prime of 0 plus phi prime of 2x by 2 should be equal to minus h prime of 0 plus phi prime of 2x plus phi prime of 0 by 2 and this happens if and only if h prime of 0 equal to 0. So similarly, u t is continuous if and only if h prime of 0 is 0. So the same compatibility condition therefore u is c 1 if and only if h prime of 0 is 0 phi of 0 equal to h of 0. Let us look at c 2s for which we need the formula for uxx in both the regions. So therefore, uxx is continuous at points of the line x equal to t if and only if phi double dash of 0 plus phi double dash of 2x by 2 is equal to h double dash of 0 plus phi double dash of 2x minus phi double dash of 0 by 2 and that is if and only if phi double dash of 0 equal to h double dash of 0. Similarly, u t t is continuous on x equal to t under the same conditions. No new compatibility conditions are required and you can easily check that you take ux and differentiate with respect to t. Similarly, take u t and differentiate with respect to x they are also continuous on x equal to t under the same conditions. In fact, note that it is enough to check for one of them because in each of the regions 1 and 2 what are the regions set of all xt such that x is less than t and set of all xt such that x is bigger than t. In each of them u of xt that we have defined is a c2 function. So therefore, uxt is same as utx in each of the regions. Therefore, the u that we have obtained is a classical solution if and only if the following compatibility conditions are satisfied. These three compatibility conditions are satisfied. This is because we do not have a psi in the problem or psi is 0. If psi was there we would have got more such conditions and this will be different actually. H prime of 0 will be in terms of psi. We have not checked the existence of ux, ut, uxx, uxt, utt at points on the line x equal to t. It is left as an exercise to you to check this using the definitions assuming these compatibility conditions which are written down on the top of this page namely phi of 0 equal to h of 0, h prime of 0 equal to 0, phi double dash of 0 equal to h double dash of 0. Now let us look at the problem 3 now. We are asked to find u of 1, 2. We have a non-homogeneous wave equation and the usual Cauchy data and Dirichlet boundary condition which is 0. This problem can be solved using many techniques. One of them is you make this 0 that is solve homogeneous problem with the same these conditions and then non-homogeneous term is handled using Duhamel principle that is one that we have already explored. Another idea is to extend this problem to whole of r. Here x is positive it is posed only for x positive extend this problem to x in r that means extend this function these functions so that you have a problem Cauchy problem for r and then you use the Dallambert formula and you get a solution. And there is a third approach sometimes we are lucky that we can spot some special solutions which satisfy this equation the non-homogeneous part. If you notice this problem we have already considered in lecture 4.10. Here we solve it again but we use parallelogram identity. So take a special solution usually we get this by inspection particularly if the right hand side are simple functions then it is easier to guess not always possible to guess but it is a trick after all. So take a special solution to the non-homogeneous wave equation we are not talking about any other conditions only equation and that v of xt is equal to 1 by 6 x square t cube plus t power 5 by 10. There could be other functions also but you have to figure out at least one function then we are on the road to solve this problem. So now consider w is equal to u minus v then w satisfies wt t minus wxx equal to 0 because both u and v solve non-homogeneous problem therefore the difference solves homogeneous equation. What is w of x0? W of x0 is ux0 minus vx0 luckily vx0 is 0 when you put t equal to 0 v of x0 is 0. So it is ux0 which we want it to be sinx similarly wt of x0 is 0 but now the problem is the boundary condition that turns out to be a non-zero function but we do not bother because we have parallelogram identity with us which will give us solution to problems like this even if the data here is h and h non-zero. Solution of the problem for w remember we want to solve w of 1, 2 we want to find. Let us draw this line and 1, 2 actually comes in this region if this is 1 unit 2 unit will be much higher somewhere here. So this is the point p we have 1 comma now let us draw the characteristic parallelogram. Of course you could stop at the 4 here or you could also go down and take this line and see where it hits. So this is q, this is r, this is s. pq lies on x minus t is equal to minus 1 therefore q is 0, 1. qr lies on x plus t is equal to 1 therefore r is 1 comma 0. ps lies on x plus t is equal to 3, rs lies on x minus t is equal to 1 therefore s is 2 comma 1. Therefore w of p is equal to w of q plus w of s minus w of r by parallelogram identity and we get w of 1 comma 2 is equal to w of 0, 1 plus w of 2, 1 minus w of 1, 0 and that is nothing but minus 1 by 60 that is the first term other things are sin 3 minus sin 1 by 2. Therefore u of 1 comma 2 is equal to w of 1 comma 2 plus v of 1 comma 2 by the definition of w because w was u minus v and that is equal to minus 1 by 60 plus sin 3 minus sin 1 by 2 plus 1 by 6 into 8 plus 32 by 10. This upon simplification becomes sin 3 minus sin 1 by 2 plus 111 by 60 this is exactly the same solution that we obtained earlier. We will summarize for a C2 function u the following equivalence was established that is u solves a homogeneous wave equation in one dimension if and only if it satisfies parallelogram identity on every characteristic parallelogram. Then we have demonstrated how the parallelogram identity is useful in solving initial boundary value problems. Thank you.