Separation of Variables for a PDE

[semithinking]

Homework Statement

Use separation of variables to find a general series solution of
u_t + 4tu = u_{xx} for 0 < x < 1, t> 0 and u(0,t) = u(1,t)=0.

Homework Equations

The Attempt at a Solution

Looking for a solution of the form u(x,t) = X(x)T(t) implies that \frac{T'}{kt} - \frac{X''}{X} = 0 \implies \frac{T'}{kT} = \frac{X''}{X} = - \lambda where \lambda is a constant.

Then we consider the following eigenvalue problem
X'' = -\lambda X for 0 < x < 1
X(0) = 0 = X(1)

If \lambda = \beta^2 > 0 then X(x) = C \cos (\beta x) + D \sin(\sin x). The boundary conditions imply that C=0 and \beta_n = (n \pi)^2 for n \in \mathbb{Z}^+. All eigenvalues are positive.

Solving \frac{T'_n}{kT_n} = -\pi^2 n^2 \implies T_n(t) = A_n e^{-k\pi^2 n^2 t}.

Therefore, u(x,t) = \sum_{n=1}^\infty A_n \sin (n\pi x) e^{-k(n\pi)^2 t} is the general series solution.

BUT! I don't believe this is correct... :P Any corrections?

 

[vela]

What did you do with the 4tu term in the original differential equation?

 

[HallsofIvy]

Homework Statement

Use separation of variables to find a general series solution of
u_t + 4tu = u_{xx} for 0 < x < 1, t> 0 and u(0,t) = u(1,t)=0.

Homework Equations

The Attempt at a Solution

Looking for a solution of the form u(x,t) = X(x)T(t) implies that \frac{T'}{kt} - \frac{X''}{X} = 0 \implies \frac{T'}{kT} = \frac{X''}{X} = - \lambda where \lambda is a constant.

No, it doesn't! Letting u= X(x)T(t) makes the equation
XT'+ 4tXT= X''T. Dividing through by XT,
\frac{T'}{T}+ 4t= \frac{X''}{X}= -\lambda
so we have
X&#039;&#039;+ \lambda X= 0[/itex]<br /> and <br /> T&amp;#039;+ 4tT= \lambda T<br /> or <br /> T&amp;#039;= (\lambda- 4t)T<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Then we consider the following eigenvalue problem <br /> X&amp;#039;&amp;#039; = -\lambda X for 0 &amp;lt; x &amp;lt; 1<br /> X(0) = 0 = X(1)<br /> <br /> If \lambda = \beta^2 &amp;gt; 0 then X(x) = C \cos (\beta x) + D \sin(\sin x). The boundary conditions imply that C=0 and \beta_n = (n \pi)^2 for n \in \mathbb{Z}^+. All eigenvalues are positive.<br /> <br /> Solving \frac{T&amp;#039;_n}{kT_n} = -\pi^2 n^2 \implies T_n(t) = A_n e^{-k\pi^2 n^2 t}.<br /> <br /> Therefore, u(x,t) = \sum_{n=1}^\infty A_n \sin (n\pi x) e^{-k(n\pi)^2 t} is the general series solution.<br /> <br /> BUT! I don&#039;t believe this is correct... :P Any corrections? </div> </div> </blockquote>