Solving the partial differential equation using the method of characteristic to find the general solution

Question

The question is solve the following partial differential equation using the method of characteristic $\frac{\partial{u}}{\partial{x}}-\frac{\partial{u}}{\partial{t}}=2t$, with the condition $u=u(x,t)$ subject to $u(x,2x+1)=e^{x}$.

My attempt:

I first calclated that $t=c-x$ where $c$ is a constant, then i rearranged to find that $c=t+x$ then let $\zeta=x$ and $\eta=t+x$, then calculated that $\frac{\partial{u}}{\partial{x}}=\frac{\partial{u}}{\partial{\zeta}}+\frac{\partial{u}}{\partial{\eta}}$ and that $\frac{\partial{u}}{\partial{t}}=\frac{\partial{u}}{\partial{\eta}}$, i then proceeded to substitute these into the pde to get that $\frac{\partial{u}}{\partial{\zeta}}=2t$, and then wrote that $\frac{\partial{u}}{\partial{\zeta}}=2(\eta+\zeta)$ since $t=\eta+\zeta$ i then intergrated to get that $u=2\eta\zeta+\zeta^{2}+f(\eta)$ which implies that $ u=2(t+x)(x)+x^{2}+k(t+x)$

Now the boundary condition given was that $u(x,2x+1)=e^{x}$ so $e^{x}=2((2x+1)+x)x+x^{2}+k(3x+1)=7x^{2}+2x+k(3x+1)$ then I let $z=3x+1$ and then tried to rearrange for $k(z)$ but i dont think this is correct, could someone please help me with this last bit? Thank you for taking the time to read through the post.

Answer 1

$$\frac{\partial{u}}{\partial{x}}-\frac{\partial{u}}{\partial{t}}=2t$$ System of characteristic differential equations : $\quad \frac{dx}{1}=\frac{dt}{-1}=\frac{du}{2t}$

First family of characteristic curves from $\quad\frac{dx}{1}=\frac{dt}{-1}\quad\to\quad x+t=c_1$

Second family of characteristic curves from $\quad\frac{dt}{-1}=\frac{du}{2t} \quad\to\quad u+t^2=c_2$

General solution of the PDE : $\quad u+t^2=F(x+t)$

$F(X)$ is any differentiable function, where $X=x+t$.

$$u(x,t)=-t^2+F(x+t)$$

Condition :

$u\left(x,t=(2x+1)\right)=e^{x}= -(2x+1)^2+F\left(x+(2x+1)\right)$

$e^{x}= -(2x+1)^2+F(X)$ with $X=3x+1 \quad\to\quad x=\frac{X-1}{3}$

$e^{\frac{X-1}{3}}= -\left(2\frac{X-1}{3}+1\right)^2+F(X) \quad\to\quad F(X)=e^{\frac{X-1}{3}}+\left(\frac{2X+1}{3}\right)^2$

So, the function $F$ is determined. Putting it into the general solution with $X=x+t$ leads to the particular solution fitting the specified condition : $u(x,t)=-t^2+e^{\frac{(x+t)-1}{3}}+\left(\frac{2(x+t)+1}{3}\right)^2$

$$u(x,t)=-t^2+e^{\frac{x+t-1}{3}}+\left(\frac{2x+2t+1}{3}\right)^2$$

Answer 2

Using the method of characteristic, we see that \begin{align} \frac{d}{dt} u(t, x(t)) = \frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}\frac{d x}{d t} = \frac{\partial u}{\partial t}-\frac{\partial u}{\partial x} = -2t \end{align} where $dx/dt = -1$. Hence we have the ode problem \begin{align} \frac{dx}{dt} = -1, \ \ \ x(2x_0+1) = x_0. \end{align} Solving the ode gives us \begin{align} x(t) = -t+C \ \ \implies&\ \ \ x(2 x_0+1) = -(2x_0+1) +C = x_0\\ \ \ \implies&\ \ \ C= 3x_0+1 \\ \implies &\ x(t) = -t+(3x_0+1). \end{align} Now, we see that \begin{align} u(t, x(t)) = -t^2+D \ \ \implies \ \ u(t, -t+3x_0+1) = -t^2+D. \end{align} Set $t= 2x_0+1$ and use the hypothesis, we get that \begin{align} u(2x_0+1, x_0) = - (2x_0+1)^2+D = e^{x_0} \ \ \implies \ \ \ D = e^{x_0} + (2x_0+1)^2. \end{align} Finally, since \begin{align} x_0 = \frac{x+t-1}{3} \end{align} then we have that \begin{align} u(t, x) = - t^2 + \exp\left( \frac{x+t-1}{3}\right)+ \left( \frac{2x+2t+1}{3}\right)^2. \end{align}