2. Use separation of variables to find ALL nonzero solutions...

2. Use separation of variables to find ALL nonzero solutions to the wave equation $u_{tt} - u_{xx} = 0$, $0 < x < \pi$, $t > 0$ satisfying each of the following set of homogeneous conditions: (a) $\begin{cases} u_x(0, t) = 0, & t > 0 \\ u_x(\pi, t) = 0, & t > 0 \\ u(x, 0) = 0, & 0 < x < \pi. \end{cases}$ (b) $\begin{cases} u(0, t) = 0, & t > 0 \\ u_x(\pi, t) = 0, & t > 0 \\ u(x, 0) = 0, & 0 < x < \pi. \end{cases}$ (c) $\begin{cases} u(0, t) = 0, & t > 0 \\ u_x(\pi, t) = 0, & t > 0 \\ u_t(x, 0) = 0, & 0 < x < \pi. \end{cases}$

Transcript

00:01 And the question we are given that u on differentiating two terms with respect to x is equal to one divided by 16 u t t.

00:11 That's it is our first equation.

00:13 And we know that u of zero comma t is equal to u of y comma t, which is equal to zero and u of x comma zero is equal to 12 sin pi x minus 6 sin of 8 pi x and u t of x comma zero is equal to zero.

00:40 This method of separation of variables is to take the solution as form of product.

00:45 So let u of x comma t is equal to a of x multiplied with b of t.

00:55 And that is a solution of particular differential equation and putting it in equation first, we get del square divided by del x square, where del is of a at a of x multiplied with b of t, which is equal to one divided by 16 del square multiplied with del square divided by del t square where del is of again a of x multiplied with b of t.

01:31 And now for a double dash, we have a double dash of x multiplied with b of t is equal to one divided by 16 multiplied with a of x multiplied with b double dash of t.

01:51 And for average, we can write a double dash of x divided by a of x is equal to b double dash of t divided by b of t multiplying with one divided by 16 from which we can write one divided by 16 multiplied with what here, double dash of x divided by a of x is equal to b double dash of t divided by b of is equal to minus lambda square.

02:33 Let lambda square is in separation constant.

02:37 Now, since 16 multiplied with a double dash of x divided by a of x is equal to minus lambda square and b double dash of t divided by b of t is equal to minus lambda square, then 16 multiplied with a double dash of x is equal to minus of lambda square a x from which we can write 16 of 16 multiplied with a double dash of x is equal to lambda square minus lambda square multiplied with a of x and we can write this as 16 multiplied with a double dash of x plus lambda square a of x is equal to zero.

03:45 And for which the auxiliary equation auxiliary equation would be 16 m square plus lambda square is equal to zero for which we get m is equal to plus minus i of lambda divided by now our general solution which means a x is equal to c1 cos of lambda divided by 4 x plus c2 sin of lambda divided by 4 multiplied with x and then we are given that u of zero comma t is equal to zero and which means a of zero multiplied with b of t is equal to zero.

04:50 So, a of zero is equal to zero or b of t is equal to zero.

04:57 We will consider only a of zero is equal to zero since b of t is equal to zero then u then u of x comma t is equal to zero which is a trivial solution but we are searching for the non -trivial solution...

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