CSIR NET MATHEMATICS | How to solve the problem of Partial Differential Equation? (June 2018)

In this video, we are going to discuss Problem of Partial Differential Equation from the CSIR NET Exam, June 2018. Let u(x,t) be the unique solution of utt - uxx = 0. And the initial condition, ux(0) is equal to f(x), ut(x,0) is equal to 0, x belongs to R, where f is a function from R to R. Satisfy the relation f of x is equal to x into 1 minus x all x belongs to 0 1 and f of x plus 1 is equal to f(x) for all x belongs to R. Then what is the value of u(1/2, 5/4)? The options are first option 1 by 8, second option 1 by 16, third option 3 by 16, fourth option 5 by 16. Answer the given partial differential equation is utt minus uxx is equal to 0, x belongs to R, t greater than 0. Now, we can rewrite this equation in the form of D dash and D operator where D operator is the partial derivative with respect to x and D dash is operator partial derivative with respect to t. Now, to generate the auxiliary equation, we replace D by m and D dash by 1. Then we have 1 minus m square is equal to 0. The roots are m equal to plus or minus 1. We can say m1 equal to 1 and m2 equal to minus 1. Therefore, the solution u(x,t) is equal to g of x plus m1t plus h of x plus m2t. Now, replacing the value of m1 and m2, we have u(x,t) is equal to g of x plus t plus h of x minus t, which is the solution of the given partial differential equation utt minus uxx is equal to 0. Now we are given the initial condition. So, we are going to use the initial condition. First initial condition, ux0 is equal to f of x. So, substitute t equal to 0 in this solution. We have g of x plus 0 plus h of x minus 0. Then we have gx plus hx is equal to fx, which is the first condition. Now, the second initial condition, utx0 is equal to 0. So, ut is the partial derivative of u with respect to t. So, we are differentiating u with respect to t. g dash x plus t minus h dash x minus t, the minus because of differentiation of x minus t with respect to t. Now evaluate at t equal to 0, we have g dash x minus h dash x is equal to 0. Now integrating both the sides, we have gx minus hx is equal to c. So we have two equations, initial condition will generate the two equations gx plus hx is equal to fx, gx minus hx is equal to c. By simple elimination method, adding equation 1 and equation 2, we have gx is equal to 1 by 2 fx plus c. And similarly, subtracting first and second equation, hx is equal to 1 by 2 fx minus c. Therefore, g of x plus t is equal to 1 by 2 f of x plus t plus c. And h of x minus t is equal to 1 by 2 f of x minus t minus c. Now the solution was u of x t is equal to g of x plus t plus h of x minus t, which has value 1 by 2 f of x plus t plus c plus 1 by 2 f of x minus t minus c. So, c by 2 plus cancel with the minus c by 2. Then, we have u x t is equal to 1 by 2 f of x plus t plus f of x minus t, which is a solution of given initial value partial differential equation problem. We have to evaluate. u at 1 by 2 5 by 4. So substitute x is equal to 1 by 2 and t equal to 5 by 4. Then we have 1 by 2 f of 1 by 2 plus 5 by 4 plus f of 1 by 2 minus 5 by 4. So by simplifying we have f of 7 by 4 plus f of minus 3 by 4. So we can rewrite f of 7 by 4 is equal to f of 1 plus 3 by 4. Then we have given the condition f of x plus 1 is equal to f x for all x belongs to R. So, therefore, f of 1 plus 3 by 4 is nothing but the f of 3 by 4. And similarly, f of minus 3 by 4 can be rewrite in the form of f of x is equal to f of x plus 1. So, you can rewrite f of minus 3 by 4 as a 1 minus 3 by 4. Therefore, f of 3 by 4 plus f of 1 minus 3 by 4 is equal to f of 1 by 4. Now, we again the value of f x when x belongs to 0, 1 close interval. So, value is x into 1 minus x. So, 3 by 4 lies between 0 and 1. So, value of f of 3 by 4 is 3 by 4, f of 3 by 4 into 1 minus 3 by 4 plus f of 1 by 4 is equal to 1 by 4 into 1 minus 1 by 4. Therefore, we have 1 by 2, 3 by 4 into 1 by 4 plus 1 by 4 into 3 by 4. Therefore, finally, we have u of 1 by 2, 5 by 4 is equal to 3 by 16 and which is the option number 3. Correct answer 3, 3 by 16. Thanks for watching. For more videos, like, subscribe, press the bell icon of NextGenMath.