 Welcome to the tutorial on one-dimensional wave equation. In this tutorial we are going to solve a few problems based on the Cauchy problem for the homogeneous one-dimensional wave equation. Recall what is the Cauchy problem? Given functions phi and psi, phi is C2, psi is C1 defined on whole of R real line. We need to solve this partial differential equation which is the wave equation utt equal to c square uxx for x in R and t positive such that it satisfies these two initial conditions ux0 equal to phi x and utx0 equal to psi x holds for every x in R. So the first problem show that solutions to Cauchy problem for the homogeneous wave equation have the following properties. What are they? If the Cauchy data consist of even functions that is the function phi and psi are even functions then solution also is even function at every time instant fixed time instant t. Similarly, if the data is consisting of odd functions solution is also odd for each fixed t and moreover if the Cauchy data is also periodic functions phi and psi are periodic functions of the same period then solutions are also periodic. So let us use the D'Alembert formula for solution to the homogeneous wave equation which is this. uxt equal to phi at these points x minus ct and x plus ct we take the arithmetic mean and then we take this mean of psi on the interval x minus ct x plus ct 1 by 2c not exactly mean it is t times mean anyway. So this is the D'Alembert formula. So phi and psi are even functions we want to show that u is also an even function. What is the meaning of even functions? Phi of minus x equal to phi x for every x in R similarly psi of minus x equal to psi x for every x in R even with respect to x equal to 0 that is the correct statement. So we want to check what u is also even function that means u of minus x comma t is equal to u of x comma t this holds only starting point is a D'Alembert formula. So let us write the formula for u of minus xt it is this now we have to use that the functions phi and psi are even functions and show that this expression indeed equal to u of xt. So now phi is even function therefore this equal to phi of x plus ct minus of this is this and here it is phi of x minus ct by 2 because we are using phi is even this is a first term then we have a second term 1 by 2c integral minus x I just write as it is from the above we will handle this separately I think it is better to work with the two terms separately. So now let us show that this integral is actually equal to x minus ct to x plus ct of psi as ds then what will happen this is precisely the expression for u xt that means we would have shown that the function u is even function. So we need to deduce this from this this is the equality of these two things. So that naturally suggests a change of variable to be used because even odd functions means some change of variable x going to minus x naturally comes into place okay let us do that. So what we have is minus x minus ct minus x plus ct this is integral that we have now we apply change of variable put psi equal to minus tau psi as ds okay here. So then what will happen inside it is psi of minus tau and ds is minus d tau okay fine and now the limits when s equal to minus x minus ct what will be tau it will be x plus ct because tau is minus s and when s is equal to minus x plus ct tau will be minus s therefore this will be x minus ct. So now we have a minus sign here that minus sign corresponds to this change of limits upper and lower limits. So I get x minus ct to x plus ct d tau minus d tau has given rise to this change of limits and psi is even therefore psi of minus tau equal to psi of tau. So therefore we have shown that u of minus x comma t we have started with a formula and we have shown it is equal to u of x t this holds for every x that means u is an even function. So let us move on to the next part if the data is odd functions phi and psi are odd functions we want to show that u is odd function odd function means what odd about 0 phi of minus x equal to minus phi x psi of minus x equal to minus psi x holds for every x in all. We want to check that u of minus x t equal to minus u of x t holds for every x in all computations are exactly similar to the earlier case. So let us start computing u of minus x comma t the Dalamert formula gives us this this and now we need to use that phi and psi are odd functions. So therefore the first thing here is equal to phi of x plus ct with a minus sign and phi of x minus ct with a minus sign. So because phi is odd function by 2 and here also we are going to use that psi is an odd function. So let us compute this integral once again use s equal to tau equal to minus s. So d tau equal to minus ds therefore the integral would become it is s equal to minus tau right. So when s is equal to this tau will be minus of that that will be x plus ct to x minus ct ds is minus d tau as before and psi of minus. But now psi is odd therefore the integral psi of minus tau is minus psi of tau. Now this minus d tau the minus here corresponds to change of this limits upper and lower and that is nothing but minus x minus ct to x plus ct psi tau d tau. Therefore what we have obtained is u of minus x comma t is equal to minus phi of x minus ct plus phi of x plus ct by 2 okay that is the first term from here this term and the second term is here 1 by 2 c is there with a minus sign and x minus c2 to this is x plus ct oh sorry this is the this is for the top one yeah this is x minus ct this is x plus ct sign okay. So this is the first term and then we have 1 minus 1 by 2 c x minus ct to x plus ct psi tau d tau which is precisely minus u of x comma t. Therefore if the data is odd solution is odd at all times. Now let us turn our attention to the last part where the functions are periodic say the period is l which means that phi of x plus l equal to phi x and psi of x plus l equal to psi x this holds for every x in r. We want to check that u of x plus l comma t is equal to u of x t. So let us start with an expression for u of x plus l comma t by the Alambert formula it is this now phi is periodic or period l therefore phi of x plus l minus ct is actually phi of x plus x minus ct. Similarly phi of x plus l plus ct is nothing but phi of x plus ct and if you look at this integral once again what is this integral it is actually on what which interval x minus ct plus l to x plus ct plus l this okay. So this under a translation by l will go to x minus ct comma x plus ct this is what we want right. So this suggests what change of variable we have to put let us put tau is equal to s minus l then what will happen d tau is ds and this integral will become when s is equal to x plus l minus ct tau will become x minus ct when s equal to x plus l plus ct tau will become x plus ct and integrand psi of s psi s is tau plus l d tau but psi is periodic of period l therefore this is equal to x minus ct to x plus ct psi of tau d tau. So therefore what we have shown is that u of x plus l comma t is equal to u of x comma t that means u is periodic. So what is the importance of this problem one if the Cauchy data is even and periodic then the solution is also even because Cauchy data is even solution is also periodic because Cauchy data is periodic. Similarly when the Cauchy data is odd and periodic then the solution is odd and periodic. If x going to u x t is even the function u is even then dou u by dou x at 0 t is 0 we will show that and if the function is odd u of 0 t is 0. So let us prove this have we proved that let us show dou u by dou x of 0 t let us compute and show that it is 0. What is it by definition limit h goes to 0 u of h comma t minus u of 0 comma 0 by h. And since we have we know that the function is even so naturally it suggests that we have to consider h positive and h negative. So we can compute the limits as h coming to h goes to 0 from positive side and negative side that is what we are going to do now. So limit h goes to 0 from the negative side let us see what it is u of h t this is exactly same difference quotient as before. Now we are going to input this information so what I do is that I put h prime is equal to minus h then what will happen h prime will become positive. So then what I have here u of minus h dash comma t minus u of 0 0 by minus h dash of course limit h goes h dash goes to 0 from positive side. This what is this I have to use that u is even function. So this is nothing but limit h dash goes to 0 from positive side is u of h dash comma t because u is even with respect to x minus u 0 0 by minus h dash this exists this limit exists. So, so what is this this is actually minus this minus can come outside minus limit h dash goes to 0 plus of u of h dash comma t minus u of 0 0 by h dash this is precisely the right hand derivative at 0 0 with respect to x this is the left hand derivative both exist and one is a negative of the other that implies that dou u by dou x is 0 both of them are 0 and dou u by dou x is 0. So, this shows that if u is an even function of x then the derivative with respect to x is 0. Now, showing this is much simpler because u is odd function right. So, u of minus x comma t the t does not play any role here you can think t is not there u equal to x t with a minus sign by odd thing. Now, let x goes to 0 then the LHS will go to u of 0 comma t right hand side will go to minus of u comma 0 t this implies that u of 0 t is 0. So, therefore, if the function is even derivative with 0 at 0 and the function is odd then the function itself is 0. What is the application of this it will be used in finding solutions to initial boundary value problems. So, let me tell you what is initial boundary value problem we have discussed this in the last lectures 4.1 we have introduced the initial boundary value problem. So, we want to solve d equal to 1 obviously and u equal to let us say homogeneous wave equation and here we prescribe u of x 0 to be phi x and ut of x 0 equal to psi x this is the Cauchy data. So, this let us say 0 and L. So, we are defining let us say u of 0 comma t is given to be 0 this is a boundary condition and u of L comma t equal to 0 this is a boundary condition these 2 are initial conditions that means we have a string finite string 0 L and we are interested in studying solutions of homogeneous wave equation in this domain which is 0 L cross 0 infinity this is where x comma t would belong to and boundary condition is given as u equal to 0 at x equal to 0 and at x equal to L for all times. Now, in finding this solution the problem will be useful. So, if you see here u of 0 t is 0, u of L t is 0. So, one would like to transform this problem to a problem which is posed on full space. So, what one would be interested in solving is this equal to 0 in R cross 0 infinity. But if you want a Cauchy problem in a full space. So, you have to give what is u x 0 that I will tell you what it is going to be some function phi tilde of x. Similarly, ut of x 0 equal to some function psi tilde of x has to be defined for R. So, that means we have to extend the functions phi and psi which are given in the interval 0 L to functions which are defined on whole of R. So, this means we have to extend these functions how do we extend so that these boundary conditions are naturally satisfied. So, what is the boundary condition u of 0 t is 0. That means this function here u of 0 t is 0. Therefore, when is u of 0 t is 0 when the function u of xt is odd function when u of xt is odd function u of 0 t is 0 we know that. Therefore, what we think is to get u of xt as a odd function we need the data to be odd function. So, this suggests x 10 phi and C as odd functions. So, this problem one actually gives us an idea of solving initial boundary value problems. So, extend them are odd function. So, that solution will be odd so that u 0 t is 0 is satisfied. But how u L t will be equal to 0 will be satisfied that we would like some kind of periodicity u of 0 t is 0 u of L t is 0 if u is periodic also. Therefore, extend this as odd functions and periodic functions of period L. Then what will happen suppose you are successfully have done this and so that the extended functions have the required smoothness namely phi tilde C 2 psi tilde C 1 we can solve this problem solution in the whole space is going to be both odd function and periodic function of period L because the Cauchy data is periodic as well as odd functions and then this naturally satisfies this condition boundary conditions are automatically satisfied if the solution in entire R of course it will be solution on 0 L as well. So, that is the way to solve. So, this is the application of the problem one. The other part where we have shown that if function is even dou u by dou x at 0 t is 0 will be applicable for a different initial boundary value problem where instead of prescribing u equal to 0 we prescribe dou u by dou u by dou x at 0 t is 0 dou u by dou x at L t is 0 then that suggests that we have to extend phi and say as even function so that solution is even and boundary conditions are satisfied 0 boundary condition is satisfied for dou u by dou x. But I also want at x equal to L therefore I will do periodically also. So, in other words what I am saying is for this problem where u is given to be phi here I will be brief it is exactly same as this u t will be psi here here instead of u 0 t or u L t what we give is dou u by dou x at L t is 0 and here dou u by dou x at 0 t is 0. In that case the suggestion is extend phi and psi as even and periodic functions of period L. So, that solution will also be even and periodic and these conditions are satisfied. Let us move to the next problem. Let u be a classical solution to the wave equation show that for each fixed y in R the new function which is defined by translating u by y is also a solution. Similarly, if u has kth derivative or kth derivative is C2 then you can also take the kth derivative with respect to x and then show that it is also a solution. And this is u of A x A t is also a solution. So, this is a translation of a solution is solution this is a dilation of a solution is solution. Let us do the first problem. So, w of x t is equal to u of x minus y comma t u satisfies u t t minus u x x equal to 0. So, now I have to compute w t and w t t w x and w x x w t is simply u t at the point x minus y comma t is better to write the arguments all the time particularly when there is a chain rule involved. So, w t t of at the point x t is u t t at the point x minus y comma t. Similarly, w x of x t is u x this y does not play any role it is just translation x minus y comma t and w x x at x t is u x x at x minus y comma t. Therefore, w t t minus w x x at any point x t is nothing but u t t minus u x x at this point x minus y comma t and that is equal to 0 because u solves the wave equation at any point x comma t and hence at point x minus y comma t also. So, that means w is a solution to the wave equation. Similarly, you can check the other two parts of this problem. So, equation is translation invariant because the coefficients are constant coefficients therefore, solution is also translation invariant. A question related to problem 2 the equation u t t minus u x x equal to 0 is invariant. This you please check under this change of variables xi equal to x and eta equal to at. The operator transforms to a square into w eta eta minus w xi xi. That means the delambition operator is not invariant under this change of variables. So, question is what are the linear transformations xi equal to a x plus b t eta equal to c x plus d t under which the delambition operator is invariant. So, that is an exercise. So, let us look at the problem 3. Let u be the solution of the Cauchy problem which is given here the wave equation here that means c equal to 3. So, u t t minus 9 u x x equal to 0 and initial condition u x 0 is this and u t is this. And you are asked to find out the value at the point 0 comma 1 by 6 and discuss the large term behavior of the solution that is what happens to u of x naught t as t goes to infinity whether the limit exists if it exists what is it at every fixed point x 0. Now, if you observe the phi x and psi x phi is discontinuous right this is in the interval minus 2 comma 2 the function is 1 outside that it is 0. So, it is a discontinuous function forget about c 2 similarly this function exactly functions both phi and psi are same and they are not continuous. So, therefore you have to exercise caution the Cauchy data does not satisfy what is required for the delambed formula to be applicable. Therefore, the formula does not apply as such because phi and psi do not satisfy the required smoothness assumptions. So, what do we mean by a solution to Cauchy problem as above when the Cauchy data is discontinuous for example as here. So, we have not studied such problems do not give up do not worry there are possibilities to interpret the delambed formula as a weak solution. So, when Cauchy data is not smooth as required towards the end of our discussion on wave equation we discussed some notions of weak solutions. So, do not worry there is some way to interpret this as a solution. So, go ahead and compute. So, such a bad Cauchy data why do we consider we consider because the computations become easier as a result it is easy to illustrate certain features of solutions that we want to show using bad Cauchy data. Of course, good Cauchy data may also be used to illustrate these features but computations become very difficult, complicated, cumbersome and our attention will be focused on actually computing solution than exhibiting the features. That is the reason why we use bad Cauchy data but we should be aware that this is not the Cauchy data for which the delambed formula is a classical solution that we should keep in mind. We have given similar reasons in the context of Berger s equation also in lecture 2.15. Let us solve this problem very easy we have to simply apply the delambed formula. So, first part was to compute U of 0 comma 1 by 6. So, we have to identify what the C is our equation is U t t minus 9 U x x equal to 0 that means C is equal to 3 because delambed formula is U of x t equal to phi of x minus C t which is 3 t plus x plus C t. So, it is x plus 3 t by 2 plus 1 by 2 into C that is 6 x minus 3 t to x plus 3 t psi s d s. Now, it is very easy U of 0 comma 1 by 6 x is 0. So, phi of 3 times 1 by 6 is minus half. See how simple the computation has become minus half to half but on this interval what is psi it is 1. So, therefore phi is also 1. So, 1 plus 1 by 2 plus 1 by 6 into this is half plus half that is 1. So, this is 1 plus 1 by 6 which is 7 by 6. So, this is the answer. Now, let us discuss the large time behavior. So, fix x naught. So, U of x naught comma t we have the formula above x naught minus 3 t plus phi of x naught plus 3 t by 2 plus 1 by 6 x naught minus 3 t x naught plus 3 t psi s d s. So, here we have 2 terms. Let us call them A and B analyze them separately. So, what happens as t goes to infinity? Observe the term A as t goes to infinity what happens? x naught minus 3 t will go to minus infinity and x naught plus 3 t will become infinity go to infinity. In particular after some time x naught minus 3 t and x naught plus 3 t both of them come out of this interval minus 2 comma 2 because they are going to minus infinity and infinity respectively they will be out of this interval after some time we can find out when they will be out. So, that phi will be 0 after some time therefore for t bigger than some number phi will be 0. So, this term will be 0 for the same reason this goes to minus infinity this goes to plus infinity but psi is defined only I mean it is non-zero only and minus 2 comma 2. Therefore, it does not matter what this interval is after some time it will be minus 2 2 2 psi s d s. So, that is going to be our large time behavior. So, since phi of x equal to 0 for all x which do not belong to minus 2 comma 2 phi of x naught minus 3 t will become 0 for t bigger than x naught plus 2 by 3. In other words, I am just looking at this I am asking when x naught minus 3 t will become less than minus 2 similarly x naught plus 3 t when will that become greater than 2 and phi of x naught plus 3 t will become 0 for t bigger than 2 minus x naught by 3. So, whenever t is bigger than both of them phi of x naught minus 3 t as well as phi of x naught plus 3 t will be 0. Therefore, for t bigger than let us call t star which is maximum of these 2 numbers phi of x naught minus 3 t equal to 0 and that will also be x naught plus 3 t that will also be 0. In other words, the term A will become 0. Now, let us look at the term B for the same reason the term B which is an integral on x naught minus 3 t comma x naught plus 3 t whenever t is bigger than the t star as above this interval is like this. This interval properly contains minus 2 comma 2. Therefore, this integration x naught minus 3 to x naught plus 3 t of psi s ds is actually minus 222 psi of s ds. So, therefore, the term B for t bigger than the t star as on the previous slide is actually 1 by 6 minus 222 psi s ds which is psi s ds is 1 that is 4 by 6 which is 2 by 3. So, therefore limit u of x naught comma t no matter what x naught is this limit exists and the limit is actually 2 by 3. Let us move to the problem 4 now. So, let phi and psi be c 2 and c 1 respectively defined on the real line such that phi and psi both are identically equal to 0 outside an interval AB that is we have an interval AB inside the interval AB they may be 0 at some points may not be 0 at some points but definitely outside that phi is 0 and psi is 0. Similarly, here phi is 0 psi is 0 and let you be a solution of the Cauchy problem for the homogeneous wave equation. Show that the function x going to u x t that means you fix a t then u f t is a function of x that is also identically equal to 0 outside an interval. In other words if the data is 0 outside interval AB solution will also be 0 outside some interval that interval may not be AB but some interval. So, the functions f g in this decomposition u x t equal to f of x minus c t plus g of x plus c t remember this is the general solution or the homogeneous wave equation we derived this in lecture 4.2. So, if the functions f and g are such that they are of compact support we will define what is a compact support f and g are compact support then the integral of psi over r must be 0 integral of psi over r makes sense and it is 0. Let us start the solution of problem 4 the starting point as before is the D'Alembert formula. Let us fix time t equal to t naught we will show that the function u of x comma t naught is 0 outside some interval. So, as before let us analyze these terms separately a and b a is a simpler one to understand. See a is definitely 0 when under 2 conditions any of the 2 conditions our a and b are here imagine x minus c t is here because x minus c t is here x minus c t will be to the right side therefore the term a is 0 because phi is 0 outside interval AB. So, what is the condition one condition x minus c t is bigger than c t naught is bigger than b that is the case when x is bigger than b plus c t naught or the other case is here x plus c t is less than a then also x minus c t is to the left side and phi is 0 on the interval x minus c t x plus c t or x plus c t is less than a c t naught. So, that means that x is less than a minus c t c t naught therefore if you are on this interval a minus c t naught b plus c t naught this is c t naught u the a is 0 a is 0 if x is not in this interval. So, if x lies outside a minus c t naught b plus c t naught phi will be 0 same reason holds for b also psi will also be 0 if you are outside this interval. So, what about b the term b same answer. So, if x lies outside this interval a minus c t naught b plus c t naught then the terms both terms a and b are 0 therefore what we can conclude is if x does not belong to a minus c t naught comma b plus c t naught that implies that u of x t is 0 u of x t may be 0 u of x t naught may be 0 even if x is there inside interval we are not making any comment on that what we are saying is u is definitely 0 if x is not in that interval. So, this is what is called data compact support implies solution compact support of course one may want to know what is this support and compact support. Let us define you can find in any book on topology or real analysis. Let us consider a continuous function from r to r because as a situation we are now currently in let us consider a continuous function then what is the definition of a support of f this is a notation support of f what we have to look at is look at those points where f is not 0 and take the closure of that. So, by definition support is a closed set by definition and if it also happens to be a compact set it is called compact support. Support is always closed by definition it is also happens to be a compact set it is called support is called compact we know in r compact sets are closed and bounded sets. Therefore, if the support is bounded then definitely it is compact. Therefore, if the function is 0 outside a bounded set then the support is always a compact support. For example, let us look at one example 0 up to 0 one after one also it is 0 in between it is something like that in this case the set where f is not 0 is 0 comma 1 that is a set and we are take the closure for the definition. So, that is a closed interval 0 1 if you notice at the end point 0 the function is 0 but it still comes under the support due to the closure that is why the end points have come in 0 and 1. Now, if you look at another function which is sin x where it is not 0 is precisely except these points where the sin is 0 which are multiples of pi. So, from r you subtract multiples of pi integer multiples of pi right and take the closure and that is r. So, the support of the sin x function or cos x function for that matter is r support of this function here like this hill is actually the interval 0 1. Of course, there are other characterization of support this is the definition of support is look at the points where the function is not 0 take the closure. The other way to look at is that support of f complement is the largest open set on which f is 0. That means take any open set on which f is 0 keep on collecting such open sets take the reunion. Therefore, we will get the largest open set on which f is 0 take the complement that will be the support of the function verify that this coincides in the previous two examples that I have given. Now, let us turn our attention to the second part. Second part said u of xt is equal to this and it is given that f and g are functions from r to r they are compact support that means they are 0 outside some bounded interval. In other words f and g are 0 outside some interval which we can take alpha beta for both of them we can take the same interval because if f is 0 outside alpha beta g is 0 outside alpha dash beta dash by taking the minimum for the left hand points maximum for the right hand points we will get common alpha beta. What is important is that f and g are 0 outside the interval alpha beta inside what happens we are not concerned right now that is given then we are asked to show that integral of psi is 0. Let us see what is phi x to start with phi x is what u of x comma 0 that is f of x plus g of x of course outside alpha beta this is 0 right. So, this is 0 outside alpha beta similarly what is psi of x that is ut of x comma 0 that is nothing but minus c f dash of x plus c g dash of x if f and g are 0 outside the interval alpha beta f dash and g dash are also 0 therefore this combination is also 0 outside alpha beta. In other words phi and psi are 0 outside alpha beta so we are in the situation of the part 1 of this problem phi and psi are 0 outside this interval therefore of course solution is of compact support for every fixed time that is fine. Now what we are interested in is the following we want to see what happens to the long term right yeah limit of u of x naught comma t as t goes to infinity we did this in a previous problem it turned out that this is going to be 1 by 2 c integral over r of psi s ds okay this is what we saw in that problem of course we computed with a specific numbers and we want to say this is 0 why because what is u of x 0 comma t you can fix a x 0 comma t what is u of x 0 comma t by this expression it is f of x 0 minus c t plus g of x 0 plus c t now as t goes to infinity x naught minus c t x naught plus c t will come out of the interval alpha beta therefore that limit is 0 therefore we conclude this that integral of psi over the support must be 0 this completes problem 4. Now let us move on to problem 5 let u be a twice continuously differentiable function which solves the wave equation u t t minus c square u x x equal to 0 prove that u is constant on a member of this family x minus c t equal to constant just write the picture just for reference let me write this so this is x minus c t equal to 0 these are the lines okay these are x minus c t equal to various constants so if the solution is constant on any one of them on this of course it will depend on x and t right but if that is not the case imagine it is constant then it will be constant on each of those lines take any other line you will be constant that will be very simple from the representation formula that we have we know that solution of the wave equation is of this form we have this formula now as x comma t this point varies on a line x minus c t equal to alpha not so let us assume that u is constant on x minus c t equal to alpha not we are going to show u is constant on any line x minus c t equal to alpha for some other for any alpha also so as x t varies on this line what happens to x minus c t or x plus c t x plus c t will vary in r entire r but we are assuming u is constant let us assume that constant is c alpha 0 so u equal to c alpha 0 on this line x minus c t equal to alpha not therefore what we have is c alpha not is equal to f of alpha not plus g of x plus c t for all the x t which are on this line that implies that g of x plus c t is equal to c alpha not minus f at alpha not it means that the function g is a constant function this implies g is a constant function proof x minus c t equal to alpha not right that is where the x t is varying that implies what c t is equal to x minus alpha not that implies g of x plus t c t equal to g of 2x minus alpha not and that is equal to this number what is that number c alpha not minus f alpha not which is a constant number it is a constant so that is that is what this implies that g of any value g is identically equal to because as x varies right if you consider this line x is varying right throughout x minus c t is what is fixed as alpha not so x varies to minus infinity infinity so therefore g is a constant function g of z equal to c alpha not minus f alpha not so it is a constant function so once it is a constant function what we have u of x t is equal to f of x minus c t plus let me call beta so let us call g is a constant function so let us call that beta real number this what we have this implies what if x minus c t is equal to some number alpha then u of x t is equal to f of alpha plus beta this that is u is constant on each of these lines x minus c t equal to alpha therefore if you have found a line on which u is constant for some alpha not u is equal to c alpha not what we are showing is on this line u of x t for x t on this line will be equal to f of alpha plus beta that means constant again thank you