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Partial Differential Equations : Separation of Variables

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Example Question #1 : Separation Of Variables

Solve for the general solution using separation of variables.

Given the following information:

$\\u_t=u_{xx} \\0\leq x\leq1 \\t>0$

$\\\text{Boundary Conditions: }\left\{\begin{matrix} u_x(0,t)=0\\ u_x(1,t)=0 \end{matrix}\right.$

$\text{Initial Conditions: }u(x,0)=cos(\pi x)$

Possible Answers:

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

$\u(x,t)=e^{-\lambda^2t}\cos(\lambda x)$

$\u(x,t)=e^{-\pi^2t}\cos(\pi x)+\sin(\pi x)$

$\u(x,t)=e^{-\pi^2t}\sin(\pi x)+\cos(\pi x)$

$\u(x,t)=e^{-\pi^2t}\cos(\pi x)$

Correct answer:

$\u(x,t)=e^{-\pi^2t}\cos(\pi x)$

Explanation:

Since the question states to use separation of variables the solution looks as follows.

$u(x,0)=cos(\pi x)$

Let

U(x,t)=T(t)X(x)

therefore the partial differential equation becomes

$\\X(x)T'(t)=T(t)X''(x) \\\\\frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=k$

is some constant therefore making the ordinary differential equation,

$\\T'(t)-kT(t)=0 \\X''(x)-kX(x)=0$

In this particular case the constant must be negative.

$\\k=-\lambda^2 \\T'(t)+\lambda^2T(t)=0 \\X''(x)+\lambda^2X(x)=0$

Solving for and results in the following.

$\\T(t)=Ae^{-\lambda^2t} \\X(x)=B\cos(\lambda x)+C\sin(\lambda x)$

From here solve for the general transient solution

$\\U(x,t)=e^{-\lambda ^2t}(A\cos(\lambda x)+B\sin(\lambda x)) \\U_x(x,t)=e^{-\lambs ^2t}(-A\lambda \sin(\lambda x)+B\lambda \cos(\lambda x))$

Lastly apply the boundary conditions to solve for the constants and in turn solve the general solution.

$\\U_x(0,t)=B\lambda e^{-\lambda ^2t}=0 \\B=0 \\u(x,t)=e^{-\pi^2t}\cos(\pi x)$

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Partial Differential Equations : Separation of Variables

Study concepts, example questions & explanations for Partial Differential Equations

All Partial Differential Equations Resources

Example Questions

Example Question #1 : Separation Of Variables

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