Question #194756

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


1
Expert's answer
2021-05-18T18:13:18-0400

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


u(x,0)=sin(3x) and u(4,t)=0,but sin(34)0u(x,0)=sin(3x) \space and \space u(4,t)=0, but \space sin(3\cdot4) \not=0

Suppose


u(x,0)=sin(3πx)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\{ u(x,t) \}=U(x,s)

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

L{d2u(x,t)dx2}=Uxx(x,s)L\Big\{ \frac {d^2 u(x,t)}{dx^2} \Big\}=U_{xx}(x,s)

Applying the Laplace transform to the equation, we have


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

The general solution can be written as


U(x,s)=Uh(x,s)+Up(x,s)U(x,s)=U_h(x,s)+U_p(x,s)

the general solution of the homogeneous equation


Uh(x,s)=C1exs+C2exsU_h(x,s)=C_1 e^{x \sqrt s}+C_2 e^{-x \sqrt s}

the particular solution of the non-homogeneous equation


Up(x,s)=C3cos(3πx)+C4sin(3πx)U_p(x,s)=C_3 cos(3\pi x)+C_4 sin(3\pi x)

d2Up(x,s)dx2=(3π)2Up(x,s)\frac {d^2 U_p(x,s)}{dx^2}=-(3\pi)^2 U_p(x,s)

substitute the particular solution into the equation


(3π)2Up(x,s)=sUp(x,s)sin(3πx)-(3\pi)^2 U_p(x,s)=sU_p(x,s)-sin(3\pi x)

((3π)2s)(C3cos(3πx)+C4sin(3πx))=sin(3πx)(-(3\pi)^2-s)(C_3 cos(3\pi x)+C_4 sin(3\pi x))=-sin(3\pi x)

((3π)2s)C3=0  C3=0(-(3\pi)^2-s)C_3=0\space \rightarrow \space C_3=0

((3π)2s)C4=1  C4=1s+9π2(-(3\pi)^2-s)C_4=-1\space \rightarrow \space C_4=\frac{1}{s+9\pi^2}

the general solution


U(x,s)=C1exs+C2exs+sin(3πx)s+9π2U(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)=0U(0,s)=0=C1+C2+0u(0,t)=0 \rightarrow U(0,s)=0=C_1 +C_2+0

u(4,t)=0U(4,s)=0=C1e4s+C2e4s+0u(4,t)=0 \rightarrow U(4,s)=0=C_1 e^{4 \sqrt s}+C_2 e^{-4 \sqrt s}+0

then


C1=C2=0C_1 =C_2 =0

So, the solution is


U(x,s)=sin(3πx)s+9π2U(x,s)=\frac{sin(3\pi x)}{s+9\pi^2}

apply the inverse Laplace transform


u(x,t)=L1{sin(3πx)s+9π2}=sin(3πx)L1{1s+9π2}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)=e9π2tsin(3πx), t0u(x,t)=e^{-9\pi^2t}sin(3\pi x), \space t\ge0

Answer: u(x,t)=e9π2tsin(3πx), t0, 0x4\space u(x,t)=e^{-9\pi^2t}sin(3\pi x), \space t\ge0, \space 0\le x\le 4


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