Answer to Question #194756 in Math for Arpan Chakraborty

Question

Using the method Laplace of separation of variables, solve du /dt=d²u/dx², u(0,t)=0, u(4,t)=0, u(x,0)=sin3x

Answer 1

the problem is posed incorrectly since the initial conditions and boundary conditions contradict each other:

$$u(x,0)=sin(3x) \space and \space u(4,t)=0, but \space sin(3\cdot4) \not=0$$

Suppose

$$u(x,0)=sin(3\pi x)$$

We apply the Laplace transform to the function u(x,t) in t, considering x as a parameter

$$L\{ u(x,t) \}=U(x,s)$$

$$L\Big\{ \frac {d u(x,t)}{dt} \Big\}=sU(x,s)-u(x,0)=sU(x,s)-sin(3\pi x)$$

$$L\Big\{ \frac {d^2 u(x,t)}{dx^2} \Big\}=U_{xx}(x,s)$$

Applying the Laplace transform to the equation, we have

$$U_{xx}(x,s)=sU(x,s)-sin(3\pi x)$$

The general solution can be written as

$$U(x,s)=U_h(x,s)+U_p(x,s)$$

the general solution of the homogeneous equation

$$U_h(x,s)=C_1 e^{x \sqrt s}+C_2 e^{-x \sqrt s}$$

the particular solution of the non-homogeneous equation

$$U_p(x,s)=C_3 cos(3\pi x)+C_4 sin(3\pi x)$$

$$\frac {d^2 U_p(x,s)}{dx^2}=-(3\pi)^2 U_p(x,s)$$

substitute the particular solution into the equation

$$-(3\pi)^2 U_p(x,s)=sU_p(x,s)-sin(3\pi x)$$

$$(-(3\pi)^2-s)(C_3 cos(3\pi x)+C_4 sin(3\pi x))=-sin(3\pi x)$$

$$(-(3\pi)^2-s)C_3=0\space \rightarrow \space C_3=0$$

$$(-(3\pi)^2-s)C_4=-1\space \rightarrow \space C_4=\frac{1}{s+9\pi^2}$$

the general solution

$$U(x,s)=C_1 e^{x \sqrt s}+C_2 e^{-x \sqrt s}+\frac{sin(3\pi x)}{s+9\pi^2}$$

from the boundary condition

$$u(0,t)=0 \rightarrow U(0,s)=0=C_1 +C_2+0$$

$$u(4,t)=0 \rightarrow U(4,s)=0=C_1 e^{4 \sqrt s}+C_2 e^{-4 \sqrt s}+0$$

then

$$C_1 =C_2 =0$$

So, the solution is

$$U(x,s)=\frac{sin(3\pi x)}{s+9\pi^2}$$

apply the inverse Laplace transform

$$u(x,t)=L^{-1} \Big\{ \frac{sin(3\pi x)}{s+9\pi^2} \Big\}=sin(3\pi x)L^{-1} \Big\{ \frac{1}{s+9\pi^2} \Big\}$$

$$u(x,t)=e^{-9\pi^2t}sin(3\pi x), \space t\ge0$$

Answer:$\space u(x,t)=e^{-9\pi^2t}sin(3\pi x), \space t\ge0, \space 0\le x\le 4$