Mixed PDE problem

Question

Can someone help me with the following problem of PDE, namely

$$u_{tt}+2u_{xt}-2u_{t}=0,u=u(x,t),u(x,0)=u_{t}(x,0)=e^{x}$$ Classify the given equation and write it in the canonical form.

We were taught in the class to identify a PDE according to the form, $$au_{xx}(x,y)+2bu_{xy}(x,y)+cu_{yy}(x,y)=d$$ where $a,b,c,d$ are functions that can depend on $x,y,u,u_{x},u_{y}.$ If $D=b^{2}-ac>0$ the equation is hyperbolic.

Here we have $D=1-0\cdot 1=1>0$ and therefore the equation is hyperbolic, and we need to introduce a substitution $y'(x)=\lambda_{1}$, but $\lambda$ is given by $\lambda_{1} = \frac{b+\sqrt{D}}{a},\lambda_{2}=\frac{b-\sqrt{D}}{a}.$ How can I apply this here since a=0 and we have $\lambda_{1}=\frac{2}{0}$?

Thank you for your time!

Answer 1

If search solution in form $$u(x,t)=v(t)e^x$$ we get answer $$u(x,t)=(t+1)e^x$$

Answer 2

Let's define new coordinates $\xi= x+at$, $\eta=t+bx$; then \begin{align} \partial_x&=\xi_x\partial_{\xi}+\eta_x\partial_{\eta} =\partial_{\xi}+b\partial_{\eta}, \tag{1} \\ \partial_t&=\xi_t\partial_{\xi}+\eta_t\partial_{\eta} =a\partial_{\xi}+\partial_{\eta}, \tag{2} \\ \end{align} hence \begin{align} u_{tt}+2u_{xt}-2u_t&=(a\partial_{\xi}+\partial_{\eta})^2u +2(\partial_{\xi}+b\partial_{\eta})(a\partial_{\xi}+\partial_{\eta})u -2(a\partial_{\xi}+\partial_{\eta})u \\ &=(a^2+2a)u_{\xi\xi}+(2a+2+2ab)u_{\xi\eta}+(1+2b)u_{\eta\eta}-2au_{\xi}-2u_{\eta} =0. \tag{3} \end{align} To put $(3)$ in the canonical form of a hyperbolic PDE, we must equate to zero the coefficients of $u_{\xi\xi}$ and $u_{\eta\eta}$, so $(a,b)=(0,-1/2)$ or $(a,b)=(-2,-1/2)$. The latter possibility must be discarded, since it leads to $\xi=x-2t=-2(t-x/2)=-2\eta$. Therefore, with $(a,b)=(0,-1/2)$, Eq. $(3)$ becomes $$ u_{\xi\eta}-u_{\eta}=0. \tag{4} $$ Now, let's solve Eq. $(4)$: $$ u_{\xi\eta}-u_{\eta}=\partial_{\eta}(u_{\xi}-u)=0 \implies u_{\xi}-u=f(\xi), \tag{5} $$ where $f$ is an arbitrary (but sufficiently well-behaved) function. Multiplying both sides of $(5)$ by $e^{-\xi}$, we have $$ \partial_{\xi}(e^{-\xi}u)=e^{-\xi}f(\xi) \implies u=e^{\xi}G(\eta)+e^{\xi}\int e^{-\xi}f(\xi)\,d\xi=:e^{\xi}G(\eta)+F(\xi), \tag{6} $$ or, in terms of $x$ and $t$, $$ u(x,t)=F(x)+e^xG(t-x/2). \tag{7} $$ To determine $F$ and $G$, we use the initial conditions: \begin{align} u(x,0)&=e^x \implies F(x)+e^xG(-x/2)=e^x, \tag{8} \\ u_t(x,0)&=e^x \implies e^xG'(-x/2)=e^x. \tag{9} \end{align} Equation $(9)$ implies $G(s)=s$; plugging this result into $(8)$, we obtain $F(x)=e^x(1+x/2)$. Finally, Eq. $(7)$ yields $$ u(x,t)=e^x(1+x/2)+e^x(t-x/2)=e^x(1+t), \tag{10} $$ which agrees with the result found by Aleksas Domarkas.