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Th fundamental solution of the heat equation in $\mathbb R^d$ is $$ E_d(x,t)=H(t) (4πt)^{-d/2} e^{-\frac{\vert x\vert^2}{4t}}, $$ so that $\Vert E_d(\cdot,t)\Vert_{L^1(\mathbb R^d)}=1,$ $ \nabla_x E_d …

Set first $u= ve^{-kt}$ so that $ \partial_t u+ku=e^{-kt}(\partial_t v-kv+kv)=e^{-kt}\partial_t v. $ The equation becomes $$ \partial_t v=\alpha\partial_x^2 v, \quad v(0,t)=0, \quad v(1,t)=M_R e^{kt}, …

I agree with the previous answer. Maybe the following observation for the global problem on $\mathbb R^n$ can help. Using the fundamental solution $H(t)(4π t)^{-n/2} e^{-{\vert x\vert^2/4t}}$ of the h …

I guess that your question is ill-formulated: you have to respect the homogeneity on both sides of the inequality. Let us say that for $f$ in the Schwartz space you always have $$ \Vert f\Vert_{W^{t,q …

Let me write an answer which is in fact too long for a comment: the heat equation itself is ill-posed locally in space. Consider the fundamental solution of $\partial _t-∆_x$ ($t$ is the time variable …

Let me try to reformulate (too long for a comment) your question by specializing it to a more particular (and more stable) case. Let $u_0:\mathbb R^2\rightarrow \mathbb R$, be a (smooth) Morse functio …

Change the variables: define $$ u(t,s,x)=v(\underbrace{\frac{t+s}{2}}_{\tau},\underbrace{{t-s}}_{\sigma},x). $$ You get $ \partial_t u+\partial_s u=\partial_{\tau} v $ and the equation becomes $$ \par …

Let us write the operator, with the notation $D_x=-i\partial_x$, $$ L=D_x^2-ik D_x+\beta,\quad\text{the equation is}\quad\partial_t\Psi+L\Psi=g. $$ We start with noticing $ \partial_t-ik D_x+\beta=e^{ …

Your last equation $$ \mathcal K=v_t-xv_z-v_{xx}=0 \tag 1$$ is indeed a particular case of Hörmander's $X_0-\sum_{1\le j\le r}X_j^2$ with $$ X_0=\partial_t-x\partial_z, X_1=\partial_x, r=1, [X_0,X_1] …