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I need to solve this problem according to these conditions where $0\leq x \leq \pi, 0\leq t$: $$u_{t}=u_{xx} \\ u(0,t)=u(\pi,t)=0 \\ u(x,0)=\sin(x)+3\sin(2x)+2\sin(5x)$$

I'm pretty sure I need separation of variables: $$u(x,t)=X(x)T(t)$$

I've been able to solve for a constant $\lambda$ such that:

$$\frac{T'}{T}=\frac{X''}{X}=\lambda \text{ where } \lambda = -\beta^{2}\\ X(x)=c_{1}\cos(\beta x)+c_{2}\sin(\beta x) \\ \Rightarrow \beta=n \\ \Rightarrow\lambda=-(n)^{2} \\ X_{n}=\sin(nx)$$ And I've gotten to solving for $T(t)$: $$ T'+ n^{2}T=0 \\ T_{n}(t)=e^{-n^{2}t} \\ X(x)T(t)=\sum_{n=1}^{\infty}e^{-n^{2}t}\sin(nx)$$

But this doesn't converge when $t=0$ so this can't be the answer. Where have I made a mistake? Thank you for your help.

IndustrialPioneer
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1 Answers1

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The mistake you do is to believe that the solution is of the form $X(x)T(t)$, which is not the case. What you have done is that you have found all solutions which can be written in this form (and they form a basis in the linear space of solutions, but you have not shown this), and they are as you found (with different notation), $$\phi_n(x,t)=e^{-n^2t}\sin(nx).$$

All that remains to be done is to find a linear combination of the $\phi_n$ which satisfies the initial condition. Since $u(x,0)$ is already given in the form of a sine series, it is a simple matter to identify the coefficients (otherwise you would have determined the sine series of $u(x,0)$ first).

Mårten W
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