Answer to Question #266776 in Differential Equations for Bhajji

Question

Determine the general solution to the equation ∂²u/∂t²=c²(∂²u/∂x²) under the boundary conditions u(0,t)=u(1,t)=0 and initial conditions u(x,0)=Φ(x), u_t(x,o)=Ψ(x)

Answer 1

Solution;

Given;

$\frac{\partial^2u}{\partial t^2}=c^2(\frac{\partial^2u}{\partial x^2}).….(1)$

Using seperation of variables look for the solution of the form;

$u(x,t)=X(x)T(t)$

Then;

$XT''-c^2X''T=0$

Divide by $c^2XT$ we get;

$\frac{T''}{c^2T}=\frac{X''}{X}=-\lambda$

Here $\lambda$ must be a constant ,so we arrive to two solutions;

(i)

$T''+c^2\lambda T=0$

Whose solution is;

$T(t)=Acos(c\sqrt{\lambda}t)+Bsin(c\sqrt{\lambda}t)$

(ii)

$X''+\lambda X=0$

Whose solution is;

$X(x)=Ccos(\sqrt{\lambda}x)+Dsin(\sqrt{\lambda}x)$

From the boundary conditions;

$u(0,t)=u(1,t)=0$

It implies that C=0

And;

$\sqrt{\lambda}=nπ\implies\lambda_n=n^2π$

n is an intenger .

For n we have the X and T solutions as;

$T_n(t)=Acos(cnπt)+Bsin(cnπt)$

And;

$X_n(x)=sin(nπx)$

Each $u_n=X_nT_n$ solves the wave equation.

The sum;

$u(x,t)=\displaystyle\sum_{n=0}^{\infin}X_n(x)T_n(t)$

By substitution;

$u(x,t)=\displaystyle\sum_{n=0}^{\infin}[A_ncos(cnπt)+B_nsin(cnπt)]sin(nπx)....(2)$

Is a solution of the equation.

Equation (2) is a solution if it satisfies the initial conditions;

If:

$u(x,0)=\Phi(x)=\displaystyle\sum_{n=0}^{\infin}X_n(x)T_n(0)=\displaystyle\sum_{n=0}^{\infin}A_nsin(nπx)$

And;

$u_t(x,0)=\Psi(x)=\displaystyle\sum_{n=0}^{\infin}X_n(x)T'_n(0)=\displaystyle\sum_{n=0}^{\infin}B_ncnπsin(nπx)$

Again,as long as $\Phi$ and $\Psi$ are piecewise continuously differentiable,then we can find $A_n$ and $B_n$ to satisfy these equations as follows;

$A_n=2\int_0^1\phi(x)sin(nπx)dx$

$B_n=\frac{2}{cnπ}\int_0^1\psi(x)sin(nπx)dx$

Hence the complete general solution to equation (1) under the given conditions is by substituting these values in (2).