Definitions — Question 9

Question 9 – Solution

Strategy. Treat the other independent variable as constant in each partial derivative. Verify the PDE before imposing the entire initial profile.

Step 1: Definitions. The independent variables are $x,t$ and the dependent variable is $u$. Written as $u_t-4u_{xx}=0$, the equation is

Step 2: Parameter relation. Direct differentiation gives $$u_t=-\lambda A e^{-\lambda t}\cos(kx),\qquad u_{xx}=-k^2A e^{-\lambda t}\cos(kx).$$ Thus the residual is $$u_t-4u_{xx}=(4k^2-\lambda)A e^{-\lambda t}\cos(kx).$$ At $x=0$ the cosine equals $1$, while $A e^{-\lambda t}\ne 0$. The residual vanishes everywhere if and only if $\boxed{\lambda=4k^2}$; no division by a possibly zero cosine is needed.

Step 3: Initial profile. At $t=0$, the condition is $A\cos(kx)=3\cos(2x)$ for every real $x$. Evaluation at $x=0$ gives $A=3$. Taking two $x$-derivatives of this identity at $0$ gives $-Ak^2=-12$, so $k^2=4$. Since $k>0$, $k=2$, and $\lambda=16$. Therefore $$\boxed{u(x,t)=3e^{-16t}\cos(2x).}$$ For this function $u_t=-48e^{-16t}\cos(2x)=4u_{xx}$, and the initial profile is recovered at $t=0$. The function is smooth for all real $x,t$; in particular it works for forward time $t\ge 0$.

Step 4: Amount of data. The profile prescribes a value for every $x$ along $t=0$. The single condition $u(0,0)=3$ fixes only $A$, leaving any $k>0$ with $\lambda=4k^2$ possible. The full profile selects one candidate in the stated family; no uniqueness assertion about all possible PDE solutions has been proved.