 we recall the last our last lecture that given a dynamic system x dot is equal to f of f function of xt ut dt ut and subject to performing index or cost function is like this way. So, first term of the cost function is a terminal cost function and second part of this one is integral cost function. Then how we have solved this problem then first we have generated or formed a Hamiltonian matrix this is the Hamiltonian matrix with the knowledge of integral of the integral cost function this one plus that lambda transpose of t of f of t this is the function of f of t behind this idea is this is the constant optimization problem is converted into a unconstrained optimization problem using what is called lag range multiplier and lag range function is now expressed in Hamiltonian function and ultimately we got some necessary condition and boundary condition to solve this problem. So, our first step in during this process our first step is del h del u assigned to a zero that is the one necessary condition then we have a del h del lambda is equal to x dot which is called the state equation this is another necessary condition and del h del x different partial differential of s with respect to x is equal to minus del lambda dot star is a cost cost state vector because dimension of this one and dimension of x and lambda are same and their expression is exactly same in place of lambda you replace by x and that x dot replace by lambda dot with preceded minus sign and this is the boundary condition we got it. So, one has to solve this equation necessary condition along with the boundary condition to get the optimal solution of our control input which in turn it will give you the optimal trajectory of the our state and hence it will minimize or maximize the what is called performing index that we have to check it with considering the sufficient condition and we have established the sufficient condition for the functional should be a cost function should be a minimize or maximize this is the matrix we have to form this matrix which if it is greater than 0 that j which is the our performing index is minimize if it is less than 0 that means less than 0 means it is a negative definite matrix and this matrix as a dimension n plus m into n plus m if it is a negative definite then j is will get it maximum that means performing index value is maximum. So, there are different situation is we have discussed first case is fixed final time and fixed final state. So, the first three equation then del h del u is equal to 0 del h del x is equal to minus lambda dot of x then lambda dot of t then del h del lambda is equal to x dot these three equation is to solve only changes is boundary conditions and since both are fixed time is fixed fixed final time is fixed and final state is fixed then our delta t f boundary condition delta t f is 0 delta x f is 0. So, there will be there is no boundary condition for that one. So, free final time this time free final time and fixed final state that means delta t f is free delta t f is free that means we have to make delta t f is free this equal to boundary condition this equal to 0. But delta x f is fixed the final time is fixed the final time final state is fixed. So, this is equal to 0. So, only we have a case b we have only one condition is the boundary condition. Similarly, final time is fixed final state is free final state is free. So, delta f is fixed means the 0. So, delta x f is free that means this boundary condition this equal to s n to 0 and f to 0 when both are free then this equal to 0 in addition to this there is another this equal to this you have to make it and solve the problem for this optimization problems. So, let us now see how to solve such type of problems in practical problems using the Hamiltonian principle formulation. So, example or example is taken some let us call we have a resistance and we have a one capacitance is there which is connected in parallel with an inductor agree. And that our problem is this is the current flowing through the circuit i of t which we denoted by u of t input to the circuit and that resistance is r and that value let us consider is a 1 and that is the inductance that value you consider that value numerical value of l is equal to 100 and c is 1 farad agree. So, let us call current flowing through this one inductor is i l of t and which we denoted the current flowing through this denoted by a variable x 1 of t agree. So, the voltage across the capacitor voltage across the capacitor with a v c of t agree that we denoted by a variable that we denoted by a variable x 2 of t this voltage across the capacitor. So, what is our problem the initial state initial state x 2 of 0 is nothing but a voltage across the capacitor which is given to 2 volt. And initial current flowing through the inductor is assumed to be this is nothing but a current flowing through the inductor is assumed to be 0. So, our problem is to find a control input u of t in other words i of t such that after a finite time interval the voltage across the capacitor and current flowing through the inductor will be 0. So, our problem is find a control law u of t such that the loss in the circuit loss power wastage in the circuit will be minimum not only this after a finite time interval t f the voltage across the capacitor and current flowing through the inductor must be 0. So, we now these are the our if you can say these are the all initial condition for the circuit or for the system. So, our problem is find the input current or input control signal u of t which is equal to i of t. So, as to waste the least possible energy in the resistor and not only this and bring the system state bring the system at a time t f finite time t f you can write finite time t f to a 0 voltage across the capacitor not only this is 0 current is 0 current through flowing the inductor through the inductor. So, now you say this is our control problem what is the we have to optimize we have to optimize the power loss in the resistor the resistor part is minimum that is we have to power and initial condition state of the initial condition of the inductor current flowing through the inductor is 0 and voltage across the capacitor is 2 we have to find a control signal i of t or u of t such that that such that the circuit attains the in a finite time the voltage across the capacitor is 0 current flowing in inductor is 0 not only that during this finite time the power wasted in the resistance part will be as minimum as possible mean minimum. So, this is our problem statement so now in order to put this problem in the framework of what is called optimization of dynamic optimization problem then we have to write the dynamic equation of the system into x dot is equal to f of x which is a function of x t u t comma t. So, let us see how we can make it that one so we know by basic what is called this voltage across this one voltage across this one is same as the voltage across the inductor is same as voltage across the capacitor. So, I can write by the l d i d t l d l current flowing through this one of t d t is equal to v l is equal to v c of t and v c is we have consider voltage across the capacitor with a variable x 2 of t and current flowing through the inductor current flowing through the inductor i l of t is a variable x 1 of t. So, we can write l x 1 dot of t is equal to x 2 of t divided by l agree since our case we have consider l is equal to 1 and d c is equal to micro 1 ferrard then we can write this is nothing but a x 2 of t. So, this value is 1 so that is let us call equation number 1 another equation we can write it if you see this one current flowing through the capacitor is nothing but a by using the k c l law culture of current law at this point i l i of t is equal to i l plus i c. So, i c current will be is equal to i c current flowing through the capacitor you can write it here is current flowing through the capacitor i c of t agree i c of t. So, i c of t i can write it total current minus current flowing through the inductor. So, you know the current flowing capacitor is equal to c d v voltage across the capacitor with respect to time is the current flowing through the capacitor is equal to i of t minus that i l of t and i of t is u of t input and i l is x 1 of t. Now, this is nothing but a v c is x 2 so it is nothing but a x 2 dot of t is equal to u of t by c minus x 1 of t by c is equal to since c is equal to 1 then it is u of t minus x 1 of t. So, that is equation number 2 so combining equation number 1 and 2 i can write x 1 dot of t x 2 dot of t is equal to this is 0 1 then it will be a minus 1 0 then is x 1 of t x 2 of t then plus 1 0 then 1 just this equation and this equation we club together this and this we club together and writing like this way agree. So, this is equal to this into u and this is a very well known structure of state space representation of a system that is we can write it this is nothing but a x dot this is x dot of t is equal to a is a matrix this matrix is a or you write a x of t the a is a constant matrix plus b u of t which we can write it is nothing but a function of x of t u of t comma t. So, it is a general form we are writing x dot is equal to function of this and this f is a 2 dimensional because it has a 2 equations are there x 1 dot x 2 dot. So, this system dynamic equation this is the system this is described into a what is called a state space form or in other words this we can represent into a second order differential equation which can be converted into a 2 first order differential equation which is written into this form. So, let us call this is equation number 3 agree. So, now what is our performing index if you see what is our performing index according to our problem that our performing index is p i performing index is you have to minimize the loss in the resistive part of the circuit that means i square into r. So, I have a that means i square into r of d t over a finite time interval t 0 to t is equal to q and t is equal to t f and this is we have to minimize but we have already discussed in our static optimization problem that if the performing index if you add something constant term or you multiply by performing index by a constant term or divide by constant term the optimal point does not change. But optimal value may change it but optimal point at which the function has a optimal value that will not change. So, we can make it this is a divided by 2 half why we made it is half we will explain this thing later because this portion because when you do the what is called our derivation everything we have shown there will be a 2 will come and 2 and half will be cancel otherwise the 2 will be carried out throughout the our formulation that is why this half. But our optimal point at which this will be optimal that will not change it but optimal value function of the optimal value may change it. So, let us see the how to solve this one. So, if you recollect our that numerical example we have considered let us consider equation number 1. So, we rewrite the equation equations as per our statement of the problem. So, we have the equation if you see x 1 dot is equal to we got it x 2 dot let us call this is 1 this set of equation at 1 x 2 dot of t is equal to minus x 1 of t plus u of t. This is equation number 1 and 2 combinedly now I am rewriting equation number what and the performing index that is what every as j is equal to t 0 to t f half this half i square i is our u t u square of t r is 1 this value r is 1. So, d t agree which this is you see r value is what so it will be that 1. So, which we can write it as x of t f the boundary condition and this is t f there is no boundary condition. So, this part practically it is 0 plus t 0 to t f t 0 is to t f t 0 is 0 or t f then our half u square d t agree. So, this quantity is now according to our problem it is nothing but a integral of the integral cost function integral. So, it is a function of x t u t and x t u t and comma t. So, this part I can write it that one. So, our problem is that if you see determine a control law or signal that will drive the system form an initial state x 1 of 0 is equal to 0 means current flowing through the inductor is 0 from this condition and current voltage across the capacitor is a 2 volt. So, this is the initial condition of the system states. So, our job is derive a control law u such that initial condition at the will come to the final state x 1 t f and x 2 t f is 0 in a finite time and as well as it will minimize this performance index. So, that is our problem so to the final state from a initial state to the final state x 1 t f is equal to x 2 t f is equal to 0 in minimum waste of energy or minimum control effort. What is the minimum control effort? Suppose you have a resistance that resistance what is that if current is flowing through the resistance what is this? Our energy involved i square r into t and in our case r is 1 so it is i square means u square. So, it is nothing but a multi input case is nothing but a u transpose u indicates the control effort. So, that control effort we have to minimize for a fixed value of resistance. So, this is the control effort to minimize or the energy wastage in the circuit is minimized this is this one second point is second point is obtain the system time response for this control law for this optimal control law. So, this you see our problem is given the system given the system minimize the performance index in such a way minimize the performance index that is this by selecting u control effort you minimize this performance. This performance index minimization means the minimum what is called waste in the loss in the resistive part will be minimized. So, we have framework this we can write it if you see this this we can write it x dot of t is equal to f of x of t u of t and t. So, we have now made it in general framework x dot is equal to this and our performance index is terminal condition in our present case terminal condition is 0 plus that integral cost function this is the integral cost function and this is the terminal cost we have framework. So, now we apply that whatever the method we have discussed earlier using the Hamiltonian function. So, this now we are in a position to solve this problem using the Hamiltonian function. So, if you see the solution and our objective is clear to bring the state from initial state to final state in a final time interval as well as it should minimize the performance index that p i and subjective that conditions. So, our solution is first you form Hamiltonian matrix Hamiltonian equation or function. So, h of this is equal to half u square of t plus lambda transpose f of x of t u of t of t and this is equal to you know half u square of t plus lambda lambda is a because the dimension of the state is 2 lambda is a lambda 1 of t lambda 2 of t multiplied by x dot of t is what you just see when you are 4 into a vector form this then we can write it this is nothing but a x 2 of t minus x 1 of t plus u of u of t this is our f of x agree this. So, if you simplify this one this into this multiply it is a half u square of t plus lambda 1 into lambda 1 into x 2 of t plus lambda 2 into this one means minus lambda 2 into x 1 plus lambda 2 of t into u of t. So, this is our Hamiltonian function. So, once we know the Hamiltonian function. So, this is our equation number one and let us call this is our equation number 2 that one equation number 2. So, on this is equation number 3 the Hamiltonian function is equation number 3. So, immediately we can find out what is called that del h del u del u of t. So, this is equal to 0 what is del h del u term is involved here and here. So, this is twice u t is equal to lambda t. So, if you see this one u of t plus lambda of t is equal to 0 u of t is equal to minus lambda 2 of t. So, once you do this one or you can write in step or first step step is one is form Hamiltonian function second step is your second step step is form del h del u third step that what is the algorithm we have written according to that we are following that it using the value of this h of this we are writing use the value of u of t replace u of t by lambda of t then it will come half this will be a half replace u of t by minus lambda 2 of t. So, this will be lambda 2 square lambda 2 t square then plus lambda 1 of t x 2 of t just h expression what we got it we have written it and replace u of t by lambda minus lambda of t only then your minus lambda 2 of t x 1 of t minus that is lambda 2 square of t. So, if you simplify this and this will be minus lambda 2 square of t other terms as it is. So, I am writing other terms as it is lambda 2 of t x 1 of t. So, this is let us call this is equation number 4 this is equation number 5 I am writing now what is the way to do it we have to find out what is called del h del lambda del h del x both the state and costate equation we can write it now. So, next is your step 4 del h del lambda is equal to del lambda means what if you see this one is nothing but a that one del h del lambda 1 del h del lambda 2 that 2 things and this I can write it del h del lambda 1 lambda 1 term is involved here is nothing but a first term this term is nothing but a x 2 del h del lambda 1 is a x 2 this is a x 2 and this will be a and this is a differential with respect to lambda 2. So, it will be a minus lambda 2 twice lambda 2 2 to cancel this lambda 2 and this will be a x 1. So, this will be a minus x 1 minus lambda 1 lambda 2 of t this one just this value I am writing and we know lambda h by lambda dot is equal to x dot that expression. So, we can write it that that is this we are differentiating with respect to lambda 2. So, this will be minus that is you see this is a minus sign agree and so this equal to what this equal to we know that will be a your x 1 dot of t x 2 dot of t look this expression del h del lambda is equal to x dot what is x dot x 1 dot is and x 2 dot I am now writing the value of delta del h is nothing but a del h lambda 1 whose value is x 2 del h del lambda whose values is minus x 1 this is the minus note this is the minus minus x 1 minus lambda 2. So, from there I can write it that one equation that if you see I can write it x 1 dot is equal to see this one this equation x 1 dot is equal to x 2 of t x 1 dot is equal to x 2 of t let us call this equation is 6 then second equation you see x 2 dot is equal to minus lambda x 1 of t minus lambda 2 of t. So, I can write it x 2 dot of t is equal to minus x 1 of t minus lambda 2 of t that is equation number 7. So, this is the state equation that now costate equation costate equation lambda h of dot x of x dot is equal to minus lambda dot of t. So, this is nothing but a del h del x 1 of t del h del x 2 of t this one is equal to minus lambda 1 dot lambda 2 dot. So, this now you differentiate this thing with respect to the del h that is what we got it here what is this expression that you differentiate this with respect to x 1 del h del x 1. So, x 1 term is there only it will be a lambda 2. So, I can write it this equal to this equal is equal to your lambda 2 because you differentiate with respect to lambda 1 lambda x 1 then if you differentiate del h del x 1 you see del h del x 1. So, if you differentiate this one it will be lambda 2 minus lambda 2. So, it is a you can write it minus dot lambda 1 is equal to minus lambda 2 of t. Therefore, lambda 1 of t dot is equal to lambda 2 of t this is. So, let us call this equation is 8. Similarly, from this equation this is I am writing from this equation this is from this equation differentiating h with respect to x 2 x 2 is involved here only x 2. So, lambda 1 of t if you differentiate with respect to x 2. So, this is a lambda 2 dot of t is equal to minus lambda 1 of t. So, let us call this is equation number 9 just this 2 equation costate equation we retain. So, this is one equation and this is another equation we got it. So, we got it costate equation now we see from equation what we can do it from equation 8 and 9 from equation 8 and 9. If you differentiate this thing with respect to t lambda 2 dot is equal to minus lambda 1 dot. So, from 8 and 9 lambda 2 dot we are differentiate once again with respect to time t is equal to minus lambda 1 dot of t. Now see lambda 1 dot of t is lambda 2 of t. So, I am writing is nothing but a minus lambda 2 of t. So, it is something like x dot is equal to minus x. So, it is a homogeneous equation the solution of this one we can get it like this way this is equal to alpha 1 e to the power of j t and alpha 2 e to the power of minus j t this one. And this if you take the Euler's formula all these things we will get alpha 1 plus alpha 2 costate plus j alpha 1 minus alpha 2 sin t. That means e to the power j t is costate plus j sin t here is costate minus j sin t. Then simplify you will get it this one let us call this equation is equation number 10. So, this is the solution of x lambda t now you see note at time t is equal to 0 lambda here lambda 2 0 is equal to alpha 1 plus alpha 2 sin t is 0. This we got it let us call this is equation number 11. Note lambda 2 dot this again from here is equal to alpha 1 plus alpha 2 alpha 1 alpha 2 is the constant which you will can solve the characteristic equation if you solve it you will get it that this roots and this is the associate with the that is the constant this is the constant real constant number may be complex also. So, this is a alpha 2 dot is equal to alpha 1 plus alpha 2 differentiation of this one is minus sin t plus j alpha 1 minus alpha 2 costate. So, at time t is equal to 0 then you can write it lambda 2 dot of 0 this value we can write it lambda 2 dot of 0 is equal to alpha 1 plus alpha 2 this is 0 into 0 plus j alpha 1 minus alpha 2 this. So, this we got it that lambda 2 is lambda dot is this is this and we know lambda dot is what lambda dot t is equal to lambda dot t is equal to lambda 2 dot t is equal to minus lambda 1 t. So, I can write it minus lambda 1 t is equal to j alpha 1 minus alpha 2 because this we know since we know from 9 if you see from equation 9 from 9 that lambda 2 dot of t is equal to 2 dot of t is equal to minus lambda 1 of t from that you got it lambda 1 0. So, this expression we got it now from 10 and 12 let us call this is the equation number is 12 from 10 to 12 from 10 to 12 what we can write it from 10 to 12 we have say 10 11 and 12 what we can get it lambda 2 of t from 10 lambda 2 of t is equal to now see this we got it this value what is lambda 1 plus lambda 2 in this value is lambda 2 of 0. So, this is equal to lambda 2 of 0 then it is cos t as it is but j alpha 1 minus alpha 2 j alpha 1 minus lambda 1 minus lambda 1 of 0. So, this is cos t minus lambda 1 of 0 sin t. So, let us call this equation number 13 and if you differentiate this one h root is equal to lambda 2 of 0 minus sin t minus lambda 1 of 0 cos t. So, this values you know minus lambda 1 of t is equal to lambda 2 of 0 minus sin t and minus lambda 1 of 0 cos t. So, let us call equation number 14. So, I know the expression for lambda 2 I know the expression for lambda 1 this is the final expression for lambda 1 and lambda 2, but still I do not know what is the lambda 2 0 all this thing if I know then our what is the langrange multiplier description is known provided if I know lambda 1 of 0 lambda 2 of 0. So, from equation 6 and 7 you see the from equation 6 and 7 I recollect this one from 6 and 7 what you can write it x double dot if you differentiate this thing with respect to once again with respect to time there is double dot is equal to x 2 dot and express the x 2 dot with this one replace x 2 dot by this one. So, I am writing now from 6 and 7 x 1 dot 6 is x 1 dot. So, I am differentiating with respect to time once again is equal to x 2 dot and x 2 dot is equal to you know minus x 1 of t plus u of t and what is u of t minus lambda of t. So, it is u of t which is nothing but a this is minus lambda of t. So, this will be a minus x 1 of t minus lambda 2 of t agree. So, what is our minus x 1 of t what is lambda 2 of t just now we have got the expression for lambda 2 of t you see the expression for lambda 2 of t we got it here. So, I will write it this expression that expression is what minus lambda 2 of 0 cos t minus lambda 1 of 0 sin t. So, that I am writing this I am writing from equation 13 agree. So, replace the lambda 2 value by this one from equation 13 this. So, this if you look at this expression here that x 1 double dot is equal to minus x 1 of t sum quantity is there. This whole thing I can consider this whole thing I can consider as an input to the systems that whole thing I can consider as if it is a input to the system or differential equation is a forcing function or it is a forcing function to the differential equation forcing function to the differential equation agree. So, now this solution you can easily find out because it is something like as double dot is equal to minus or you bring it this one x 1 double dot is equal to plus x 1 of t plus some forcing function. So, that solution one can find out easily that one. So, the let us call the solution of x 1 of t therefore x 1 of t is equal to twice sin t. So, I leave this thing as an exercise to check the solution of that one is this that that second order differential equation what we got it lambda 1 of t sin of t minus t cos of t minus half lambda 2 of 0 t sin of t. So, we have done up to equation 14 this equation let us call this is the equation of 15. So, our control law is nothing but u of t is nothing but a minus lambda 2 of t. So, lambda 2 of t if you know this one we can find out the control law once we find out the control law then we can find out the trajectory x of t. But here lambda 2 of 0 lambda 2 of lambda 1 of 0 is unknown this is unknown this is unknown. So, you have to find out these values then only lambda 2 is the once lambda 2 is there then we can find out the control law. So, our trajectory is this one still if you see that lambda 1 0 lambda 2 0 is unknown to us and not only this that this another thing you see x 2 this is our x 1 t and we know from the dynamic equation x 1 dot of t is equal to x 2 of t that is called from 16 this is from dynamic equation of the system equation of the system see equation number 1. So, x 1 dot if you differentiate this one x 1 dot is equal to x 2 which equal to this right hand side if you differentiate this one twice cos t plus half lambda 1 of 0 this one will be what sin cos t minus t this one you have to differentiate that thing with respect to this. So, first is differentiate this one let us call this will be then cos t then unchanged minus sin t minus sin t then this part minus half lambda 2 of 0 then sin t then plus t cos t see just I have differentiate this one and that is nothing but our x 2. So, our x 2 expression therefore our x 2 expression x 2 expression is nothing but a twice cos t twice cos t plus half lambda 1 0 then if you simplify that one then what will come just I am simplifying these things then lambda 1 t half lambda 1 0 t sin t minus half lambda 2 0 sin t because this and this are cancel. So, it will be a minus this is minus minus plus. So, it will be half lambda 2 0 t sin t and this is as it is this part will be this part will be as it is here reproduced sin t plus t cos t now this trajectory I mention is u of t is from the condition we got minus lambda of lambda 2 of t and lambda 2 of t is a function of if you say is a function of that our lambda 1 of 0 lambda 2 of 0 again once I know lambda 1 0 lambda 2 0 then lambda t is known here also you see x 1 t is a function we need the information of lambda 1 of 0 and lambda 2 of 0. Then we can find out x t trajectory optional trajectory x 2 trajectory which will control input as well as x 1 t x 2 to combine it will minimize the our performing index that consider. So, applying now what is our left with us only boundary condition from the boundary condition we have to find out x 1 lambda 1 of 0 lambda 2 of 0 boundary condition and in the boundary condition our t f is free what is fixed that x t f is fixed means both the state that current flowing to the inductor current voltage across the capacitor must go to the 0. So, applying now applying the given end point boundary condition we get at t is equal to t f x 1 t f current flowing in the inductor this is x 2 t f current voltage across the capacitor is 0 this is the we know the our boundary condition. So, if you apply the now boundary conditions this one if you recollect this our boundary condition boundary condition h del s x of t this delta t whole thing put t is equal to t f delta t f plus del s x t of t del x of t minus lambda of t whole t is equal to t f transpose this transpose just see the boundary condition multiplied by this multiplied by that is I am writing multiplied by your delta x f is equal to 0 now this delta x f if you see that delta x f is in our case is 0. So, only this is not equal to 0 so our condition is delta dot and this since we do not have any terminal condition this part is 0 no terminal condition according to our problem statement. So, this you find out t is equal to t f this is equal to 0. So, this is our boundary condition that finally boils down to this equations. So, from equation 5 what is the equation 5 you just see that our basic Hamiltonian equation Hamilton function this equation from equation 5 I will write it t is equal to t f. So, from equation 5 from 5 put t is equal to from 5 t is equal to t f then your half minus lambda square of t f agree that is half lambda square of t f this is what this equation you see half lambda square of t f then lambda 1 t f x 2 t f minus lambda 2 t f then x 1 t f is equal to 0. So, what I did it here you see h of this put the value t is equal to t f is equal to 0. So, this equation is 0 because terminal what is called final state variable values is 0 this is 0. So, only this one so this one will lambda 2 t f is equal to 0. So, from 13 you see equation number 13 refer to equation number 13 that one lambda 2 of t is this one agree. So, from equation 13 t is equal to t f. So, lambda 2 t f is equal to lambda 0 lambda 2 0 see this equation lambda t is equal to t f this lambda 2 0 cos t minus lambda 1 0 sin t f t is equal to t f is t f this quantity is 0 just we got it this is 0 t is 0. So, it is a we can say this is lambda 2 0 cos t f minus lambda 1 0 sin t f this is equal to 0 that is agree. So, let us call this is equation number what is the equation number we got last equation number 9 16 then you give it this equation number is 17 agree this x 2 expression that x 2 expression what we got it x 2 twice cos of sin that is you give it 17 because this will refer 17 this you give it 18 agree. Now see this one this is unknown this is unknown and t f is unknown agree 3 unknowns are there we need 3 equation which 3 equation are your. So, I can put it now that this 18 from 15 to 15 and 17 t is equal to t f at t is equal to t f 15 is that one you see 15 and 17 is that one 15 and 17 15 is this one 17 is that that one in this expression t is equal to t f you put it t is equal to t f t is equal to t f. So, left hand side is 0 left hand side 0. So, you have a 3 unknowns are there and 3 equations are there, but that 3 equations are non-linear. So, you have to solve it to get the final value of lambda of 0 lambda 1 of 0 lambda 2 of 0 and t f and that will give you the total picture of what is called what is the optimal control law as well as what is the optimal trajectory which will minimize the performance index of that one. So, a part of this one is left. So, I will continue next class agree this one. So, I will stop it here now.