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Solve the partial differential equation $$u_t=3u_{xx}, u(0,t)=0, u(x,0)=\cos{x}\sin{5x}$$ Attempt: Using separation of variables, let $u(x,t)=f(x)g(t)$, so $$f(x)g'(t)=3f''(x)g(t)$$ $$\frac{g'(t)}{3g(t)}=\frac{f''(x)}{f(x)}=-\alpha$$ Each individual equation can be solved into $$g(t)=C_0e^{-3\alpha t}$$ $$f(x)=C_1\cos{x\sqrt{\alpha}}+C_2\sin{x\sqrt{\alpha}}$$ $$u(x,t)=f(x)g(t)$$ So since $u(0,t)=f(0)g(t)=C_0C_1e^{-3\alpha t}=0$, for the function to not be identically $0$, $C_1$ must be $0$. So now we have $$u(x,t)=Ce^{-3\alpha t}\sin{x\sqrt{\alpha}}$$ But now I am stuck, as I cannot find a way to possibly satisfy the second initial condition $u(x,0)=\cos{x}\sin{5x}$.

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Let $u(x,t)=f(x)g(t)$ ,

Then $f(x)g'(t)=3f''(x)g(t)$

$\dfrac{g'(t)}{3g(t)}=\dfrac{f''(x)}{f(x)}=-\alpha^2$

$\begin{cases}\dfrac{g'(t)}{g(t)}=-3\alpha^2\\f''(x)+\alpha^2f(x)=0\end{cases}$

$\begin{cases}g(t)=c_3(\alpha)e^{-3t\alpha^2}\\f(x)=\begin{cases}c_1(\alpha)\sin x\alpha+c_2(\alpha)\cos x\alpha&\text{when}~\alpha\neq0\\c_1x+c_2&\text{when}~\alpha=0\end{cases}\end{cases}$

$\therefore u(x,t)=\int_0^\infty C_1(\alpha)e^{-3t\alpha^2}\sin x\alpha~d\alpha+\int_0^\infty C_2(\alpha)e^{-3t\alpha^2}\cos x\alpha~d\alpha$

$u(0,t)=0$ :

$\int_0^\infty C_2(\alpha)e^{-3t\alpha^2}~d\alpha=0$

$C_2(\alpha)=0$

$\therefore u(x,t)=\int_0^\infty C_1(\alpha)e^{-3t\alpha^2}\sin x\alpha~d\alpha$

$u(x,0)=\cos x\sin5x=\dfrac{\sin6x+\sin4x}{2}$ :

$\int_0^\infty C_1(\alpha)\sin x\alpha~d\alpha=\dfrac{\sin6x+\sin4x}{2}$

$C_1(\alpha)=\dfrac{\delta(\alpha-6)+\delta(\alpha-4)}{2}$

$\therefore u(x,t)=\int_0^\infty\dfrac{\delta(\alpha-6)+\delta(\alpha-4)}{2}e^{-3t\alpha^2}\sin x\alpha~d\alpha=\dfrac{e^{-108t}\sin6x+e^{-48t}\sin4x}{2}$

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  • $\begingroup$ Thanks, it was that pesky product to sum trig identity I was missing. $\endgroup$ – ant11 Mar 18 '14 at 0:07
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$u_t=3u_{xx},\; u(0,t)=0,\; u(x,0)=\cos{x}\sin{5x} \qquad (1)$

$\cos{(x)} \sin{\left( 5 x\right) }=\frac{\sin{\left( 4 x\right) }}{2}+\frac{\sin{\left( 6 x\right) }}{2}$

Let $u_1$ is solution of

$u_t=3u_{xx},\; u(0,t)=0,\; u(x,0)=\frac{\sin{\left( 4 x\right) }}{2} \qquad (2)$

$u_2$ solution of

$u_t=3u_{xx},\; u(0,t)=0,\; u(x,0)=\frac{\sin{\left( 6 x\right) }}{2}. \qquad (3)$

Then $u=u_1+u_2$ is solution of $(1)$.

We seek for $u_1$ in the form $u = v(t)\sin(4x)$. Substituting this expression in $(2)$ we obtain

$v'(t)=-48v(t),\; v(0)=\frac12$. $\Rightarrow$ $v(t)=\frac12 e^{-48t}$, $u_1=\frac12 e^{-48t}\sin(4x)$.

We seek for $u_2$ in the form $u = v(t)\sin(6x)$. Substituting this expression in $(3)$ we obtain

$v'(t)=-108v(t),\; v(0)=\frac12$. $\Rightarrow$ $v(t)=\frac12 e^{-108t}$, $u_2=\frac12 e^{-108t}\sin(6x)$.

Finally,

$$u=u_1+u_2=\frac12 e^{-48t}\sin(4x)+\frac12 e^{-108t}\sin(6x)$$

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