Boundary Value Problem$ 5U_x+2U_y=(x-y)^2$

Question

Solve the following boundary value problem

$5U_x+2U_y=(x-y)^2$

$U(3y,y)=\exp(y^3)$

Hint: Use change of variable $(x,y)\rightarrow(e,n)$

\begin{align}e &= 2x - 5y\\ n &= x - y\end{align}

I am struggling to understand why those boundaries have been picked, they do not seem to simplify the expression in any way. Usually one of the terms will cancel. Am I missing something?

Answer 1

The hint is correct. Such change variables is really helpful: $$ \begin{cases} U_x=2U_e+U_n\\ U_y=-5U_e-U_n \end{cases} \quad\Rightarrow\quad 3U_n=n^2 \quad\Rightarrow\quad U=\frac{n^3}{9}+f(e) \quad\Rightarrow\quad U(x,y)=f(2x-5y)+\frac{(x-y)^3}{9} $$ Using the condition $U|_{x=3y}=e^{y^3}$ you can find function $f(y)=e^{y^3}-\frac{8y^3}{9}$, whence follows $$ U(x,y)=e^{(2x-5y)^3}-\frac{8(2x-5y)^3}{9}+\frac{(x-y)^3}{9}\,. $$

Remark. To make the change of variables $$ \begin{cases} e=2x-5y\\ n=x-y \end{cases} $$ notice that $$ \begin{align} U(x,y)=U\bigl(x(e,n),y(e,n)\bigr)=\widetilde{U}(e,n)\\ \widetilde{U}(e,n)=\widetilde{U}(e(x,y),n(x,y))=U(x,y) \end{align} $$ Hence follows $$ \begin{cases} U_x=\widetilde{U}_e e_x+\widetilde{U}_n n_x=2\widetilde{U}_e+\widetilde{U}_n\\ U_y=\widetilde{U}_e e_y+\widetilde{U}_n n_y=-5\widetilde{U}_e-\widetilde{U}_n \end{cases} $$ Formally, the tilde mark is needed. But two-and-a-half-century MathPhysics tradition insists on omitting any tildes or their substitutes.

Answer 2

Hint: This is a general approach called the method of characteristics.

Inverting the variable transformation you have

$$x = (5n-e)/3, \\\ y = (2n-e)/3$$

Applying the change of variables let

$$\hat{U}(n,e)=U[x(n,e),y(n,e)]$$

Then

$$\hat{U}_n= U_x\frac{\partial x}{\partial n}+U_y\frac{\partial y}{\partial n}=\frac{1}{3}(5U_x+2Uy) = \frac{1}{3}n^2$$

This can be integrated easily to find the general solution -- then apply the initial condition

Answer 3

You shall not pay attention to the hint but applying the so called method of characteristics for the present hyperbolic linear equation, which can be written in the form:

$$f(x,y) \, u_x + g(x,y)\, u_y = h(x,y),$$ where $f$ and $g$ are constants and $h = (x-y)^2$. Then, $u$ is a constant over some curves called the characteristics, which are to be obtained from the equations:

$$ \frac{dx}{f} = \frac{dy}{g} = \frac{du}{h}. $$ The first equality tells us that, after substitution of your data, $2 dx - 5 dy = 0$, therefore:

$$2x - 5y = \eta,$$ for some constant, $\eta$. To solve for $u$, you may take one of the two remaining equalities. Either $dx/f = du/h$ or $dy/g = du/h$. For example:

$$ \frac{du}{(x-y)^2} = 5^2 \frac{du}{(3x-\eta)^2} = \frac{dx}{5}, $$

which yields a separable equation for $u = u(x;\eta)$, whose solution is given by:

$$u(x;\eta)= \alpha (3x-\eta)^3+ \beta, $$ with $\alpha$ a known value, while $\beta$ can be put as a function of $\eta$, so we would have the solution, substituting the value of $\eta$:

$$u(x,y) = \bar{\alpha} (x+5y)^3 + \beta(2x - 5y),$$ where $\bar{\alpha}$ is another known constant. If my math is correct, you just have to substitute your condition $u(3y,y) = e^{y^3}$ in order to obtain the unknown function $\beta$.

Cheers!