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Find formal solution to Initial Boundary Value Problem. {partial u} / {partial t} = 3 {partial^2...

Question:

Find formal solution to Initial Boundary Value Problem.

{eq}\frac{\partial u}{\partial t}=3\frac{\partial^2 u}{\partial x^2}+x,0<x<\pi,t>0, {/eq}

{eq}u(0,t)=0,u(\pi,t)=0,t>0, {/eq}

{eq}u(x,0)=\sin x,0<x<\pi {/eq}

Formal Solution For Initial Boundary Condition:

The formal solution for initial boundary condition is given as the sum of particular and the complimentary solution for the given equation to find the complimentary solution first the auxiliary equation is found and with respect to the complimentary solution the particular solution is found.

Answer and Explanation:

Writing complimentary equation for the given function

{eq}\frac{\delta U}{\delta t}=D\\ \frac{\delta^2 U}{\delta x^2}=D' {/eq}

{eq}(D^2-3D'^2)U=x {/eq}

The function will be equated to zero

{eq}(D^2-3D'^2)U=0\\ DU=3D'U\\ \frac{\delta U}{\delta t}=3\frac{\delta^2 U}{\delta x^2} {/eq}

Using separation of variable

Let

U=X(x)T(t)

{eq}U_t=X(x)T'(t)\\ U_{xx}=X''T {/eq}

So,

{eq}XT'=3X''T\\ \frac{T'}{T}=\frac{3X''}{X}\\ \frac{T'}{3T}=\frac{X''}{X}=constant (k)\\ So,\\ \frac{T'}{3T}=k\\ T'-3kT=0\\ (D-3k)T=0\\ m=3k\\ T=C_1e^{3kt} {/eq}

Also

{eq}\frac{X''}{X}=k\\ X''=kX\\ (D'^2-k)X=0\\ m^2-k=0\\ m^2=k\\ m=\pm k\\ X=C_2e^{kx}+C_3e^{-kx} {/eq}

But when the constant (k ) is negative

{eq}T=C_1e^{-3kt}\\ and\\ X=C_4\cos kx+C_5\sin kx\\ {/eq}

When k = 0

{eq}T=C_1'\\ And X=(C_6+C_7)x {/eq}

Among all the solutions from the value of k the solution which is the best is

{eq}U=TX\\ =C_1e^{-3x}(C_4\cos kx+C_5 \sin kx)\\ {/eq}

So complimentary solution is

{eq}U_{complimentary}=e^{-3kt}(C_4'\cos kx+C_3'\sin kx) {/eq}

Particular solution for this complimentary will be

{eq}U_{particular}=\frac{1}{D-3D'^2}x\\ =\frac{1}{D(1-\frac{3D'^2}{D})}x\\ =\frac{1}{D}(1-\frac{3D'^2}{D})x\\ =\frac{1}{D}(x-0)\\ =xt {/eq}

Summing up both the solution

{eq}U=e^{-3kt}(C_4'\cos kx+C_5' \sin kx)+xt {/eq}

Applying given boundary conditions

{eq}U(0,t)=e^{-3kt}(C_4'(1)+C_5'(0))+0=0\\ C_4'e^{-3kt}=0\\ C_4'=0\\ So\\ U=e^{-3kt}(C_5'\sin kx)+xt=4\\ U(x,0)=(1)C_5'\sin kx+0=\sin x\\ C_5'\sin kx=\sin x\\ C_5'=1\\ k=1\\ U=e^{-3(1)t}\sin x +xt\\ U=e^{-3t}\sin x+xt {/eq}


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