Given the following inhomogeneous heat equation, how to show that the solution is unique?:
$\frac{\partial v(x,t)}{\partial t}- K\frac{\partial^2 v(x,t)}{\partial x^2}=-c{v(x,t)}+y(x)$
$ x\in[0,L]$
IC: $ v(0,0) = v(L,0) = 0 $ and is twice cont. diff.
BC: $ v(0,t) = v(L,t) = 0 $
The constants c, K and L are all $> 0$
$y$ is cont. on $[0,L]$
I'm wondering if these kind of PDE's can rewritten as:
$ \frac{\partial v(x,t)}{\partial t}- K\frac{\partial^2 v(x,t)}{\partial x^2}= k$
The last is one of which i guess can be solved by showing that only 1 unique solution exists by introducing a new function $w$ which is equal to the difference of 2 solutions $w_1 - w_2$. Since these are equal the k cancels out. Writing it down:
$ \frac{\partial w_1(x,t)}{\partial t}- K\frac{\partial^2 w_1(x,t)}{\partial x^2}= k$
$ \frac{\partial w_2(x,t)}{\partial t}- K\frac{\partial^2 w_2(x,t)}{\partial x^2}= k$
So $w = w_1 - w_2$ which yields the homogeneous equation $\frac{\partial w(x,t)}{\partial t}- K\frac{\partial^2 w(x,t)}{\partial x^2}= 0$
Your equation can be rewritten using the transformation $\tilde{v}(x,t)={\rm e}^{ct}v(x,t)$. Then, \begin{align} \frac{\partial}{\partial t} \tilde{v}(x,t)=&{\rm e}^{ct}\frac{\partial}{\partial t}v(x,t)+c{\rm e}^{ct}v(x,t)\\ =&K\frac{\partial^2}{\partial x^2}\tilde{v}(x,t)+{\rm e}^{ct}y(x) \end{align} Without this transformation if you had two solutions you can take the difference $w=v_1-v_2$ and $w$ would solve homogeneous equation: \begin{align} \frac{\partial}{\partial t} w(x,t)=K\frac{\partial^2}{\partial x^2}w(x,t)-cw(x,t) \end{align} A good method to prove uniqueness is the energy method where you consider \begin{align} \frac{\partial}{\partial t}\int_0^L\frac{1}{2}w(x,t)^2dx=&\int_0^L w(x,t)\frac{\partial}{\partial t}w(x,t)dx\\ =&\int_0^L w(x,t)\left(K\frac{\partial^2}{\partial x^2}w(x,t)-cw(x,t)\right)dx\\ =&-\int_0^L\left( K\left[\frac{\partial}{\partial x}w(x,t)\right]^2+cw(x,t)^2\right)dx\leq 0 \end{align} Thus the $L^2$ norm of $w$ is not increasing so if it begins as zero it will remain zero, which implies that $w(x,t)=0$. Importantly, you need the correct initial conditions $w(x,0)=0$ for $x\in [0,L]$.