Summary of Separation of Variables — Question 6

Question 6 – Solution

Strategy. Changing coordinates changes the differential operator and the integration measure; a further amplitude factor restores ordinary sine orthogonality.

Step 1: Transform the operator, not just the variable. The chain rule gives $u_x=U_s/x$ and $(x^2u_x)_x=U_{ss}+U_s$. With $U=e^{-s/2}V$, direct differentiation yields $$\boxed{V_t=V_{ss}-\tfrac 14V,\qquad 0<s<1,\quad V(0,t)=V(1,t)=0.}$$ Ignoring $U_s$ would shift every eigenvalue incorrectly.

Step 2: Recover the spectrum and field. The sine modes give $$X_n(x)=x^{-1/2}\sin(n\pi\log x),\qquad \lambda_n=(n\pi)^2+\tfrac 14.$$ Since $dx=e^sds$, the two $x^{-1/2}$ factors cancel the measure in a product: $$\int_1^eX_mX_n\,dx=\int_0^1\sin(m\pi s)\sin(n\pi s)\,ds.$$ Thus the squared norms are $1/2$, with zero cross products. The solution is $$\boxed{u=e^{-\lambda_1t}X_1+2e^{-\lambda_2t}X_2.}$$

Step 3: Use the correct projection. For initial data $f$, $$\boxed{b_n=2\int_1^ef(x)x^{-1/2}\sin(n\pi\log x)\,dx =2\int_0^1e^{s/2}f(e^s)\sin(n\pi s)\,ds.}$$ It is the transformed field $V(s,0)=e^{s/2}f(e^s)$ that has ordinary sine coefficients. For example, the ungauged sines are not orthogonal in $dx$: $$\int_1^e\sin(\pi\log x)\sin(2\pi\log x)\,dx =-\frac{4\pi^2(e+1)}{(1+\pi^2)(1+9\pi^2)}\ne 0.$$

Step 4: Verify diffusion and dissipation. Each $X_n$ satisfies $-(x^2X_n')'=\lambda_nX_n$ and vanishes at both endpoints. The temporal factors therefore verify the original PDE and exactly recover the supplied initial coefficients. Integration by parts gives $\tfrac 12(d/dt)\int u^2=[ux^2u_x]_1^e-\int x^2u_x^2=-\int x^2u_x^2$. On a mode $a e^{-\lambda_nt}X_n$, both sides equal $-\lambda_na^2e^{-2\lambda_nt}/2$, since $\int x^2X_n'^2=\lambda_n\int X_n^2=\lambda_n/2$.