How would we solve the same initial value problem but instead of c∈R being a constant we have a function f:Rn×[0,t)→R in its place such that the problem becomes:{ut+b⋅(Dxu)+f(x,t)u=h(x,t)in Rn×(0,∞)u(x,0)=g(x)on Rn×{t=0}What steps would I take to find a function u(x,t) that satisfies this? I am assuming u(x,t) will be similar to the function I found for the original problem but with some additional integrals (after playing with it for a little), but I am unsure.I am fairly new to partial differential equations so any help will be appreciated.

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How would we solve the same initial value problem but instead of c∈R being a constant we have a function f:Rn×[0,t)→R in its place such that the problem becomes:{ut+b⋅(Dxu)+f(x,t)u=h(x,t)in Rn×(0,∞)u(x,0)=g(x)on Rn×{t=0}What steps would I take to find a function u(x,t) that satisfies this? I am assuming u(x,t) will be similar to the function I found for the original problem but with some additional integrals (after playing with it for a little), but I am unsure.I am fairly new to partial differential equations so any help will be appreciated.

Marquis Ibarra · Accepted Answer

Step 1I&#039;ll illustrate with&amp;nbsp;n=2, but of course this process is true for any dimension. I&#039;ll also adopt the t,x convention, i.e writing&amp;nbsp;u(t,x1,x2). Writing the equation in typical form, and giving it some general initial data∂tu+b1∂1u+b2∂2u=h-f&amp;nbsp;uu(γ(r1,r2))=φ(r1,r2)Where&amp;nbsp;γ:ℝ2→ℝ3&amp;nbsp;is a parametric surface in (t,x) space (parameterized by, in this case, I am using r as the parameters). In our example of courseγ(r1,r2)=(0,r1,r2)Step 2We write our characteristic equations, where I have introduced the notations&amp;nbsp;x=(x1,x2),b=(b1,b2)&amp;nbsp;and&amp;nbsp;r=(r1,r2)&amp;nbsp;to save space∂st(r,s)=1∂sx(r,s)=b∂sv(r,s)=h(t(r,s),x(r,s))-f(t(r,s),x(r,s))&amp;nbsp;v(r,s)Coupled with the initial datat(r,0)=0x(r,0)=rv(r,0)=φ(r)I have chosen v as the function so that, once we have inverted these expressions to find r(t,x),s(t,x) thatu(t,x)=v(r(t,x),s(t,x))Your equation in your comment is not quite right. Forgetting some rigor when it comes to arguments, it should read roughlyv′(s)=h(x,t)−f(x,t)v(s)Not just&amp;nbsp;v′(s)=−f(x,t)v(s)

How would we solve the same initial value

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