MHB Why is the general solution of this form?

[evinda]

Hello! (Wave)I found the following in my lecture notes:$$u_t=u_{xx}, x \in \mathbb{R}, t>0 \\ u(x,0)=f(x)$$

$$u(x,t)=X(x)T(t)$$

$$\Rightarrow \frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda \in \mathbb{R}$$

$$X''(x)+\lambda X(x)=0, x \in \mathbb{R}$$
$$X \text{ bounded }$$

The characteristic equation is $\rho^2+ \lambda=0 \Rightarrow \rho^2=-\lambda$

• $\lambda<0$: $X(x)=c_1e^{ \sqrt{- \lambda}x}+c_2 e^{- \sqrt{- \lambda}x} $
  
  So that $X$ is bounded $\Rightarrow c_1=c_2=0$.
  
• $\lambda=0$: $X(x)=c_1x+c_2 \overset{c_2=0}{\Rightarrow } X_0(x)=1$
• $\lambda>0 \Rightarrow \rho^2= \lambda i^2 \Rightarrow \rho= \pm i \sqrt{\lambda}$
  
  $\Rightarrow X(x)=c_1 \cos(\sqrt{\lambda}x)+ c_2 \sin (\sqrt{\lambda} x), x \in \mathbb{R}$
  
  $T'(t)+ \lambda T(t)=0$
  
  $\Rightarrow T(t)=c e^{- \lambda t}$
  
  $$u_{\lambda}(x,t)=e^{- \lambda t} \cos(\sqrt{\lambda} x) \\ e^{- \lambda t} \sin(\sqrt{\lambda}x)$$
  
  General solution:
  
  $$u_1(x,t)= \int_0^{\infty} c_1(\lambda) e^{- \lambda t} \cos(\sqrt{\lambda}x)d \lambda$$
  
  $$u_2(x,t)= \int_0^{\infty} c_2(\lambda) e^{- \lambda t} \sin(\sqrt{\lambda}x)d \lambda$$
  
  $$u(x,t)= \int_0^{\infty} c(\lambda) e^{- \lambda t} e^{i \sqrt{\lambda} x} d \lambda$$
  
  $$( c_1, c_2: [0,+\infty) \to \mathbb{R}, c:[0,+\infty) \to \mathbb{C}$$
  
  Initial conditions:$$u(x,0)=f(x) \Leftrightarrow f(x)= \int_0^{\infty} c(\lambda) e^{i \sqrt{\lambda}x} d \lambda$$
  Could you explain me why in this case the general solution is of the form $u(x,t)= \int_0^{\infty} c(\lambda) e^{- \lambda t} e^{i \sqrt{\lambda} x} d \lambda$?
  
  Why isn't it $u(x,t)= \frac{a_0}{2}+\sum_{k=1}^{\infty} (a_k \cos (kx)+ b_k \cos (kx))$ ?
  

 

[I like Serena]

Hey! (Smile)

Why isn't it $u(x,t)= \frac{a_0}{2}+\sum_{k=1}^{\infty} (a_k \cos (kx)+ b_k \cos (kx))$ ?

I think you're only taking $X(x)$ into account, while the solution is $X(x)T(t)$. (Wasntme)Let's look at it like this. (Thinking)

We found that for any $\lambda > 0$:
$$u(x,t)=X(x)T(t) = (c_1 \cos(\sqrt{\lambda}x)+ c_2 \sin (\sqrt{\lambda} x)) c e^{- \lambda t}
=C e^{-\lambda t} \cos(\sqrt{\lambda}x) + D e^{- \lambda t} \sin (\sqrt{\lambda} x)$$
is a solution.

Any linear combination with different $\lambda$'s will also be a solution.
So a more general solution is:
$$u(x,t) = \sum_i \Big[C_i e^{-\lambda_i t} \cos(\sqrt{\lambda_i}x) + D_i e^{- \lambda_i t} \sin (\sqrt{\lambda_i} x)\Big]$$
where $\lambda_i$ is any positive number, and $C_i,D_i$ are any arbitrary constants.

Letting $i$ go to infinity, we can write this as:
$$u(x,t) = \int_0^\infty \Big[C(\lambda) e^{-\lambda t} \cos(\sqrt{\lambda}x) + D(\lambda) e^{- \lambda t} \sin (\sqrt{\lambda} x)\Big]d\lambda$$
(Mmm)

Next step is to apply Euler's formula. (Thinking)